Prove or disprove: into a cube box with edge 4, one can put 65 balls of diameter 1.

The volume of a ball with diameter 1 is  nearly 0.5235. The volume of a corresponding cube with a radius 1 is 1 and the difference is nearly 0.476. It is obvious that we can put into a cube with edge 4 64 balls of a radius 1. When some balls are touching each other or walls of a cube, no matter how we place them, we'll never get a free continious volume higher then nearly 0.476. So there are no place for 65th ball.

reminds me of the second FMO-final question, but this is in three dimensions :)

i suppose it is possible...

[hide]
first layer from the bottom: 16 balls with their midpoints in a perfect square grid
second layer: place 9 balls with their midpoints above the center of four balls of the first layer. Pythagoras shows us that the distance between the plane containing the midpoints of the first layer and the plane containig the midpoints of the second layer, is only $\frac{\sqrt{2}}{2}$ instead of $1$
third layer: again 16 balls
fourth layer: again 9 balls

the distance from the bottom of the box till the planes with the midpoints of the nth layer, are $\frac{1}{2}, \frac{1}{2}+\frac{\sqrt2}{2}, \frac{1}{2}+\sqrt{2}, \frac{1}{2}+\frac{3\sqrt2}{2}$
This means we still have a space of approximatly $1.378679656$ left above our fourth layer. So we can still add a layer of 16 balls.
so if you arrange them nicely, $3*16+2*9=66$ balls fit in the box :) [/hide]


geometry
3D geometry
sphere
search
advanced fields
advanced fields unsolved

Show that every closed curve $ c$ of length less than $ 2\pi$ on the surface of a unit sphere lies entirely on the surface of some hemisphere of the unit sphere.


number theory
relatively prime
superior algebra
superior algebra unsolved

Let $r,s,t$ positive integers which are relatively prime and $a,b \in G$, $G$ a commutative multiplicative group with unit element $e$, and $a^r=b^s=(ab)^t=e$.
(a) Prove that $a=b=e$.
(b) Does the same hold for a non-commutative group $G$?

for the last part :
consider G=S5
a=(12345)
b=(23)(45)
r=5
s=2
t=3

you really need commutativity

For a)
$(ab)^{rt}=b^{rt}=e=b^s \Rightarrow b=e$ since $(rt,s)=1$
$a=e$ with same reasonment


the lengths of the three sides of triangle ABC are consecutive integers. if AB = x^2-12x-64, BC = x+5, and CA = x+3, find the perimeter of ABC

[hide]Clearly $BC$ is 2 more than $CA$, so the order of the sides from least length to greatest is $CA,AB,BC$. Now we have that $(x+3)+1=x^{2}-12x-64$
$x^{2}-13x-68$
$(x-17)(x+4)$
$x=17 ,-4$. Since $x=-4$ gives a negative side length, we clearly must discard this solution, so $x=17$.
Plugging in $x=17$ gives lengths of 20, 21, and 22, so the perimeter is $20+21+22=\boxed{63}$[/hide]


[b]TurkZeka 2005/2 Puzzle Competition wil start at 14 May 2005  22:00:00 GMT. [/b]

This year, each competition (total 4 times in a year-this one is second competition) is going to have 10 puzzles and these puzzle competitions are open to all country.

Competitions will be published in only [b]Turkish, English and Japanese[/b].

You will have exactly 7 days to submit the answers of the puzzles via "Answer" form page, beginning from the puzzle's publishing date.

You should read information about answer format in puzzle text carefully before answering. Each correct answer is 100 points and additionally there are two types of  bonus points.

One of the bonus points are given to the first 3 competitors who give the correct answer (30, 20 and 10 point) and the other bonus points are given  depending on answering day, beginning from current puzzle's publishing date (30, 25, 20, 15, 10 and 5 point).

Correct answers and solutions will be given when the next puzzle is published. Your score will be updated when you post your answer. You can see whether your answer is correct or not on the "Score Anlayze Table". To compete, you have to sign up to the site and log in.

Please tell friends who are interested in puzzles about these competitions and the site.

Our web address : http://www.turkzeka.com

----------------------------------------------------------------------------------------------------------------------

Aim : The site TurkZeka.Com was established with a purpose of helping puzzle fans improve their ability of imaging, thinking, finding winding key, heading conclusion, finding out the hidden properties etc. and having nice and useful time at the same time.

Nonograms :

Total 5 new nonograms are published every day.

Nonograms' sizes : 10x10 - 15x15 - 20x20 - 25x25 - 30x30

New types of puzzles will posted coming soon. (Diamond, Mines...)

Just Solve & Enjoy...

 ------------------------------------------------------------------------------------------------------------

Yilmaz Ekici

(Owner & Administrator of TURKZEKA)

Our Editors :

ISMAIL TURK : Project & Software

OZTURK YILDIRIM : Category Puzzles

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 e-mail : turkzeka@turkzeka.com

Haha...sounds interesting. It also sounds like it should be in another board. :P

Yeah, it does look cool, but I don't know if you are supposed to uhh... advertise...


This is an ugly attempt to a solution:

[b]Step 1.[/b]
Denote $x=\frac{b}{a}$ and $y=\frac{c}{a}$. Then the LHS becoms
$F(x,y)=\frac{1}{\sqrt{1+3x^{4}+3y^{4}}}+k\left(\frac{x^{4}}{y}+\frac{y^{4}}{x}\right)$.
[b]Step 2.[/b]
We prove that $F(x,y)\geq F\left(\sqrt[4]{\frac{x^{4}+y^{4}}{2}},\sqrt[4]{\frac{x^{4}+y^{4}}{2}}\right)$.
Indeed, under such transformation $\frac{1}{\sqrt{1+3x^{4}+3y^{4}}}$ remains constant, so it remains to prove that $\frac{x^{4}}{y}+\frac{y^{4}}{x}\geq\frac{x^{4}+y^{4}}{\sqrt[4]{\frac{x^{4}+y^{4}}{2}}}$, or $\frac{x^{5}+y^{5}}{x^{4}+y^{4}}\geq\frac{xy}{\sqrt[4]{\frac{x^{4}+y^{4}}{2}}}$.
It can be easily shown that $\frac{x^{5}+y^{5}}{x^{4}+y^{4}}\geq\frac{x+y}{2}$. Using $\sqrt[4]{\frac{x^{4}+y^{4}}{2}}\geq\frac{x+y}{2}$, it remains to prove $\left(\frac{x+y}{2}\right)^{2}\geq xy$, which is evident.
[b]Step 3[/b]
So we are investigating the single variable inequality:
$f(x)=\frac{1}{\sqrt{1+6x^{4}}}+2kx^{3}\geq\frac{2\sqrt{2}}{3}$.
Denote $g=1+6x^{4}$. Let $x_{0}$ be a minimum point of $f$ on $[0,\infty)$. Then $ f'(x_{0})=0$. Since $ f'=6kx^{2}+(g^{-\frac{1}{2}})'=6kx^{2}-\frac{1}{2}g^{-\frac{3}{2}}\cdot g'=6kx^{2}-\frac{1}{2}g^{-\frac{3}{2}}\cdot24x^{3}=6x^{2}\left(k-\frac{2x}{(\sqrt{g})^{3}}\right)$. Since $ f'(x_{0})=0$, we have $k=\frac{2x_{0}}{(\sqrt{g})^{3}}$.
Because we must have $f(x_{0})=\frac{2\sqrt{2}}{3}$, then $\frac{1}{\sqrt{g}}+\frac{4x^{4}}{(\sqrt{g})^{3}}=\frac{2\sqrt{2}}{3}$. It easy to see that $x^{4}=\frac{1}{6}$ satisfies the last equation [And it is the only solution!]. From $k=\frac{2x}{(\sqrt{g})^{3}}$, we obtain $k=\frac{1}{6^{4}\sqrt{2}}$.

The fact that $x^{4}=\frac{1}{6}$ is the only solution can be proved after setting $z=6x^{4}$ in $\frac{1}{\sqrt{g}}+\frac{4x^{4}}{(\sqrt{g})^{3}}=\frac{2\sqrt{2}}{3}$. We will get by multiplying everything that $9(1+z)^{2}+4z^{2}+12z(1+z)=8(1+z)^{3}$, or $(z-1)(8z^{2}+7z+1)=0$ with the only positive solution $z=1$, or $x^{4}=\frac{1}{6}$.


[hide="Hint"]
Since Bob has a 100% chance of getting the first question right, you only need to worry about him getting 2, 3, or 4 questions right out of questions 2-5.
[/hide]

[quote="i_like_pie"]Bob is taking a test with 5 questions. Bob has a 100% chance of getting the first question right, an 80% chance of getting the second one right, 60% for the third, 40% for the fourth and 20% for the last question. What is the probability that Bob gets at least 3 questions correct?[/quote]
[hide]
he will definately get the first question right.
the chance of getting the second and third and fourth wrong is 6/125. the chance of getting the second and third and fifth wrong is 8/125. the chance of getting the second, fourth, and fifth wrong is 12/125. the chance of getting the third, 4th, and 5th wron gis 24/125. added together, it is 50/125. the chance of getting 2nd, 3rd, 4th, and 5th wrong is 24/625. it has already been counted 4 times, so it is subtracted 3 times. 250-72=178. 625-178 is 447. so the chance of not missing 2 is 423/625. simplified, it is $\frac{447}{625}$
[/hide]

[quote="i_like_pie"]Bob is taking a test with 5 questions. Bob has a 100% chance of getting the first question right, an 80% chance of getting the second one right, 60% for the third, 40% for the fourth and 20% for the last question. What is the probability that Bob gets at least 3 questions correct?[/quote]

[hide]
We can just do case work for the compliment.

The probability of him getting none correct out of the other four is: $\frac{1 \cdot 2 \cdot 3 \cdot 4}{5^{4}}=\frac{24}{625}$.

The probability of him getting only one correct out of the other four:
Question 2: $\frac{4 \cdot 2 \cdot 3 \cdot 4}{5^{4}}$
Question 3: $\frac{1 \cdot 3 \cdot 3 \cdot 4}{5^{4}}$
Question 4: $\frac{1 \cdot 2 \cdot 2 \cdot 4}{5^{4}}$
Question 5: $\frac{ 1 \cdot 2 \cdot 3 \cdot 1}{5^{4}}$

The sum of both cases is $\frac{178}{625}$, and thus the desired probability is $\boxed{\frac{447}{625}}$.[/hide]


Can anyone figure this out?

Some new cookies started appearing in The Bakery very recently. These cookies are in the shape of different letters of the alphabet. The Breadmaster started laying the cookies out to cool, and you observed the letters as he laid them out in this order:

H B N A I E P R O

What are the next 3 letters he will lay out?

Bah, people probably you couldn't get this was it was from anther site, heres a hint:

[hide]This problem is from http://www.neopets.com[/hide]

I will post the answer since this question was really confusing :

[hide]H B N A I E P R O =
H  B    N

A   I    E

P   R   O

"P"" T" "P"

What is being attempted to spell out here is Happy Birthday Neopets![/hide]


advanced fields
advanced fields unsolved

I desperately need help, cause I cannot find anything about hermite cubic elements and I need to make fortran program who solves this equation with hermite cubic elements:

[img]http://cache.imagefap.com/images/full/39/182/1825031998.jpg[/img]

please explain how to solve this problem
need answer fast for semestral signature

thank you very much

Im very sorry my image did not displayed

[img]http://student.fizika.org/~sjelicic/picture001.JPG[/img]

please help me I dont know what to do

[quote="Mr.Doe"]Wa5p prijatelju, iskreno sumnjam da netko ovdje govori hrvatski ( a tvoj tekst jest na hrvatskom ). Mogao si barem to prepisati fino na engleski, i lijepo objasniti sta su to Hermiteovi kubicni elementi, gdje si tocno zapeo, pa bih ti netko mozda izasao u susret. Ovako nazalost, nista nece biti od toga... Inace koji to profesori sada dijele potpis iz kolegija  :wink: . Mozda bi cak bilo bolje da si to post-ao na DeGiorgiju, ili na forumu fizike, tamo bi ti mozda netko prije odgovorio.

POzdrav[/quote]

Što se tiče DeGiorgija, a pogotovo foruma fizike imam osobne razmjerice sa tim ljudima tako da ih izbjegavam.

Dodatno, ja i ti nismo prijatelji!


trigonometry
calculus
derivative
geometry
circumcircle
geometry proposed

Let $ ABC$ be a triangle for which $ 0 < A \minus{} C < 90^{\circ}$. Prove that

$ (a \plus{} c)\cdot\cos C\le \frac {3R\sqrt 3}{2}$, where $ R$ is the length of the circumradius.

We have equality iff $ \frac a2 \equal{} \frac {b}{\sqrt 3} \equal{} \frac c1\ .$

[b]Particular case.[/b] $ A \equal{} 90^{\circ}\ : \ h \plus{} \max \{b,c\}\le\frac {3a\sqrt 3}{4}$,

where $ h$ is the distance of the vertex $ A$ to the opposite side.

See http://mateforum.ro/viewtopic.php?p=595#595

$ (a \plus{} c)\cdot\cos C\le \frac {3R\sqrt 3}{2}\Leftrightarrow (sinA \plus{} sinC).cosC\leq \frac {3\sqrt 3}{4}$
We have $ (sinA \plus{} sinC).cosC\leq (1 \plus{} sinC).cosC \equal{} cosC \plus{} \frac {sin2C}{2}$
Let $ f(C) \equal{} cosC \plus{} \frac {sin2C}{2}$ with $ 0 < C < \frac {\pi}{2}$ we have:
$ f'(C) \equal{} \minus{} sinC \plus{} cos2C \equal{} \minus{} 2sin^2C \minus{} sinC \plus{} 1 \equal{} (sinC \plus{} 1)( \minus{} 2sinC \plus{} 1)$
$ f'(C) \equal{} 0\Leftrightarrow sinC \equal{} \frac12\Leftrightarrow C \equal{} \frac {\pi}{6}$
$ \Rightarrow max f(C) \equal{} f(\frac {\pi}{6}) \equal{} \frac {3\sqrt 3}{4}$

Please if is possibly, without derivatives and give a geometrical interpretation !

[quote="Virgil Nicula"][color=darkred]Let $ ABC$ be a triangle for which $ C < A$. Prove that $ (a \plus{} c)\cdot\cos C\le \frac {3R\sqrt 3}{2}$,
where $ R$ is the length of the circumradius. Is equality iff $ \frac a2 \equal{} \frac {b}{\sqrt 3} \equal{} \frac c1\ .$[/color][/quote]
[b]Proof.[/b] Observe that $ C < A\implies C < \frac {\pi}{2}$, i.e. $ 2C < \pi\ .$

Thus, $ \left\|\begin{array}{c} \boxed {(a \plus{} c)\cdot\cos C\le \frac {3R\sqrt 3}{2}} \\
 \\
\left(\begin{array}{c} a \equal{} 2R\sin A \\
\ c \equal{} 2R\sin C\end{array}\right)\end{array}\right\|$ $ \Longleftrightarrow$ $ (\sin A \plus{} \sin C)\cos C\le \frac {3\sqrt 3}{4}$ $ \Longleftrightarrow$

$ 2\sin C\cos C \plus{} 2\sin A\cos C\le \frac {3\sqrt 3}{2}$ $ \Longleftrightarrow$

$ \sin 2C \plus{} \sin (A \plus{} C) \plus{} \sin (A \minus{} C)\le \frac {3\sqrt 3}{2}$ $ \Longleftrightarrow$

$ \sin 2C \plus{} \sin B \plus{} \sin (A \minus{} C)\le\frac {3\sqrt 3}{2}$, what is truly

because $ \{2C,B,A \minus{} C\}\subset (0,\pi )$ and  $ 2C \plus{} B \plus{} (A \minus{} c) \equal{} \pi\ .$

Is equality iff  $ 2C \equal{} B \equal{} A \minus{} C \equal{} \frac {\pi}{3}$, i.e.  $ A \equal{} \frac {\pi}{2}$, $ B \equal{} \frac {\pi}{3}$, $ C \equal{} \frac {\pi}{6}$,

what means $ \frac a2 \equal{} \frac {b}{\sqrt 3} \equal{} \frac c1\ .$

[b]Remark.[/b] I used the remarkable relations in any triangle $ ABC$ : $ \sum\sin A \equal{} \boxed {\frac pR\le\frac {3\sqrt 3}{2}}\ .$
Denote the point  $ D$ of the circumcircle of the triangle $ ABC$ for which  $ \widehat {BAD}\equiv \widehat {BCD}$.

Observe that in $ \triangle DAC$ we have $ (\widehat {DAC}) \equal{} A \minus{} C$, $ m(\widehat {ADC}) \equal{} B$, $ m(\widehat {ACD}) \equal{} 2C\ !$


The sum of the first $ N$ positive integers is 2006.  Wait, that's impossible$ \ldots$ one of those integers must not have been included!  If in fact exactly one of the first $ N$ positive integers was not included in the sum, which integer must that have been?

the sum of the first 63 integers is 2016, so the answer is 10.

The equation of the sum of the first n numbers is n(n+1)/2.
Plugging 2006 in gives us:
n(n+1)/2 = 2006
n(n+1)   = 4012
We can then estimate n by finding the square root of 4012 (I'm assuming this was a target round problem), which is about 63.3.
Then we plug 63 back into the previous formula and get
63 * 64/2 = 2016.  Therefore the answer is 10.


I just want to complete what kennyworm wrote
if x is real we consider an open interval in the  neighbourhood of x : (x-1/n,x+1/n)
where n>0
so there exists rn belongs to Q/ rn belongs to (x-1/n,x+1/n)
so r1 belongs to (x-1,x+1)
r2 belongs to (x-2,x+2)............
then rn is a sequence of rational numbers such that rn-x belongs (-1/n,1/n)
so -1/n<rn-x<1/n
then 0<lim(rn-x)as n tends to infinty<0 (sandwish theo)
then lim rn=x
bec rn is rational we can say that f(rn)=k.rn
if f is continous as n tends to infinity f(rn)=f(x)
lim f(rn)=lim f(k.rn)=k.lim(rn)=kx
so f(x)=kx

this obviously belongs to some other post.  Perhaps a moderator could help out?


Alice sells an item at $\$10$ less than the list price and receives $10\%$ of her selling price as her commission. Bob sells the same item at $\$20$ less than the list price and receives $20\%$ of his selling price as his commission. If they both get the same commission, then the list price is

$ \textbf{(A)}\ \$20\qquad\textbf{(B)}\ \$30\qquad\textbf{(C)}\ \$50\qquad\textbf{(D)}\ \$70\qquad\textbf{(E)}\ \$100 $

Hmm, I just saw this exact problem on a local competition only without multiple choice.

[hide]
B. $\$30$

Let P = list price.
$0.1(P-10)=0.2(P-20)$
$P-10=2P-40$
$P=30$
[/hide]

[hide="Answer"]$0.1\cdot (x-10)=0.2\cdot (x-20)\Rightarrow 0.1x-1=0.2x-4\Rightarrow x=30\Rightarrow \boxed{B}$[/hide]


Hi, I am stuck.

Suppose I have a polynomial with variables like this:

$ f(x, k, n) \equal{} (x \plus{} k)^n \plus{} 3k^{n \minus{} 1} x \minus{} k^n$

I want to show that for a given $ n$ and $ k$, if I know that the roots of $ f(x,k,n)$ are $ x \equal{} \{\lambda_1,\lambda_2,...\}$, then the roots of $ f(x,\alpha k,n)$ are $ x \equal{} \{\alpha\lambda_1,\alpha\lambda_2,...\}$.

I could prove it directly for n=2 and n=3, because I can compute the roots and just compare them. But I am stuck starting n=4. Any idea on how to proceed?

- Thanks

Have you tried actually computing $ f(\alpha \lambda_i, \alpha k, n)$?

Hi, it is blowing up my mind. Let me just talk out loud to think about the meaning.

Given f with a fixed k and n, I can assume that the roots are $ \{\lambda_i\}$. Now, I claim that if k is changed to $ \alpha k$, the roots will be $ \{\alpha\lambda_i\}$. If this is true, then when I compute $ f(\alpha\lambda_i,\alpha k,n)$, I will get 0. So if I plug in $ \alpha\lambda_i$ and get 0, then I have proven the relation.

The part that blew my mind was I don't know what $ \lambda_i$ are, how would I cancel the terms to show that $ f(\alpha\lambda_i,\alpha k,n)$ goes to 0?

But let me try it and see what happens.

What you do know is that $ f(\lambda_i, k, n) \equal{} 0$ by definition.  So if you could write $ f(\alpha \lambda_i, \alpha k, n)$ in terms of $ f(\lambda_i, k, n)$...

It works! Thanks.

If f was like these (without one of the x) it wouldn't have worked:

$ f(x, k, n) \equal{} (x\plus{}k)^{n}\plus{}3k^{n\minus{}1}\minus{}k^{n}$
$ f(x, k, n) \equal{} (x\plus{}k)^{n}\plus{}3k^{n\minus{}1}x^2\minus{}k^{n}$

It worked because the power of k and x always sum up to be the same for each term.


A parcel of land was purchased for $8000. If the value of hte land increases by 15% each year, what is the minimum number of years it will take for the value of the land to double?

If you work it out, it will take 4 years for the land's amount of money to double.  In the end it will be 16,160.81775 dollars.  Long decimals, huh?

I don't think that's right, 4 years gives $8000*1.15^{4}=13992.05$.  5 years (which is the answer) will give 16090.8575.

Yes. It is indeed 5 years?

Is there a way to do it without a calculator or massive calculations¿

no i dont think there is a way. this is a target, so it doesn't really matter...
and bpms... i think he is allowed to post because he is clearly citing the source of the problems, its not as if he's saying these are his own problems.

if you want a quick answer on countdown, just divide 72 by the percent, then round up to the nearest integer, and you have your answer

try it if you don't believe me

i have no idea why this works

jorian

[quote="davidlizeng"]i don't get it... what do you mean??
it makes no sense[/quote]
Actually, the "magic" 72 does make sense, but it works the best for 5-14% (since we are rounding to the nearest integer, it doesn't really matter, though).  davidlizeng, if you wanted to get a good [b]estimate[/b] of what the value of n would be for  $(1+\frac{x}{100})^{n}$ to become 2; where 5 is about less than or equal to x and x is about less than or equal to 14, then n is about equal to 72/x.

anyways, it's gotten to a point where it's plain annoying

i hope ur parents aren't giving u a happy meal toy cause u got 2500 posts!

jorian

Dude, like no? I just posted them because I wanted to check my answers

[quote="#H34N1"][quote="bpms"]I don't think you can post all of these problems since mathcounts might have copyright laws protecting the tests.[/quote]

Its called Fair Use.[/quote]
There's a reason mathcounts isn't in the contest archive.


a.) Determine - with proof - the largest cube that fits into a unit tesseract (hypercube, 4-dimensional cube).

b.) Generalization: Determine - with proof - the largest n-dimensional cube that fits into a unit (n+1)-dimensional cube.

I wonder what is answer for $n=2$  :?  :)

Assume that ABCDEFGH is a unit cube with A(0|0|0), B(1|0|0), C(1|1|0), D(0|1|0), E(0|0|1), F (1|0|1), G(1|1|1) and H(0|1|1). It can be proven easily that WXYZ with W(1|0,25|0), X(0,25|1|0), Y(0|0,75|1) and Z(0,75|0|1) is the largest square (A = 9/8) that fits into the unit cube.

Is there anyone who can solve this problem?!

You just need find someone who is interested in this problem...

This sum is very difficult :( . can someone solve it?'

Sincerely,


there is cyclic in the addition of prime fielde (modulo p...) but what about multiplication? is there any law in their order?
I know the numbers 1....p-1 must appear, each one once, but what is their order?
I pay attention that in the half upper numbers, the law is that in the beginning, from the numbers 1...p-1, if this row is for example 3, than the numbers=0 mod 3, after that= 1 mod 3 etc, and from small to big.
for example: $\mathbb{Z}_{7}$ \[\left( \begin{array}{ccccccc}* & 1 & 2 & 3 & 4 & 5 & 6 \\ 1 & 1 & 3 & 3 & 4 & 5 & 6 \\ 2 & 2 & 4 & 6 & 1 & 3 & 5 \\ 3 & 3 & 6 & 2 & 5 & 1 & 4 \\ 4 & 4 & 1 & 5 & 2 & 6 & 3 \\ 5 & 5 & 3 & 1 & 6 & 4 & 2 \\ 6 & 6 & 5 & 4 & 3 & 2 & 1 \end{array}\right)\]

I dont quite see what your trying to get at.

I don't know if there's any law for this, but $(ax+b)(cx+d) \equiv bd \text{ (mod x)}$   ...

It seems like you're looking at the binary structure $<Z_{7}, *>$, where * denotes multiplication modulo $7$. (a [i]binary structure[/i] is a set with an operation, call it *, such that if $a$ and $b$ are unique elements of the set, then $a$*$b$ returns a unique element in the set). This binary structure is not a [i]group[/i] (which is a special type of binary structure that has a unique identity, has a unique inverse for every element, and has an associative binary operation) if that's what you're thinking about. There is a similar structure that is a group, namely,$<U_{7},*>$, where $U_{7}$ denotes the set of all positive integers less than $7$ and relatively prime to $7$. A pattern that occurs in the [i]Cayley Table[/i] (the matrix/table you posted is a Cayley Table) of $<U_{7},*>$ is that each row and each column is a permutation of the elements of $U_{7}$. This pattern follows from the three properties that define a group.

The properties of the group $<U_{7},*>$ are much nicer than the properties of the binary structure that you are looking at. I'm not sure of too many nice patterns in the table that you've listed.


Hello everyone,

These are incredibly easy questions but I do not seem to be able to solve them at night! I will appreciate any hints and help.

Thanks a lot.

---

1. Determine all integer values of x so that y will always be an integer in the given relations.

a) $ y \equal{} 3\plus{}\frac{5}{x\minus{}2}$

2. For each of the following, determine all integer points (x, y) on its graph. ([i]Hint:[/i] In part a, note that x + 4 = (x - 2) + 6)

a) $ y \equal{}\frac{x\plus{}4}{x\minus{}2}$

a) $ \frac{5}{x\minus{}2}$ has to be an integer, [b]catalan[/b]. It means that 5 has to be divisible by x-2. Therefore, $ \vert x\minus{}2\vert \equal{} 5$ or $ \vert x\minus{}2\vert \equal{} 1.$

You need to proceed likewise in b).  :D

Answer to 1
[hide]For y to be a integer, 5/(x-2) have to be integers. so therefore 5/(x-2) is an integer. So |x-2|=5 & |x-2|=1
To get x, use x-2=5 and x-2=-5 and x-2=1 x-2=-1. [hide]so therefore x={-3, 1, 3, 7}[/hide][/hide]
[/hide]

Sorry, I oopsed. See edit.

is number 2 possible? i set y to 1 and mulitplied both sides by $ x\minus{}2$ and got 6=0

Catalan, how did you get(Hint: In part a, note that x + 4 = (x - 2) + 6)?

Isn't it x + 4 = (x - 2)y? like abacadaea had mentioned?

he/she says you need to find the intercepts on the graph i think, so the x intercepts are -4 and 2...i think

Hello,

Thank you very much for your replies. Could anyone explain why absolute value would be use for 1a ?

@DonkeyKong: The hint was given in the textbook.

Thanks again.

[quote="catalan"]@DonkeyKong: The hint was given in the textbook.

Thanks again.[/quote]

Oh.. Sorry I doubted you and your textbook...


The Mock AMC that was just held by JSRosen was really good.  It was fun and good practice.  I am thinking that maybe the AoPS classes could include something of this sort.  The AMC problem series classes would probably be good classes to include them in.

My other suggestion is to hold some sort of contest.  The winner(s) of this contest would then get to enroll in one of your classes for free(or something like that).

These are just my suggestions.

What are the instuctors' thoughts on my suggestions?  Don't worry I wont be hurt if you don't like 'em ;)

Rep's comment points to the problem right now-> time!  It's something we'll put on the list of 'good ideas to think more about when there's time'.

Thanks for the suggestion.


Line segment $BC$ with length $a$ is a (fixed) chord in circle $O$. Take point $A$ on circle $O$. Let $h$, $k$ and $l$ be the lengths of the perpendicular lines in $\triangle ABC$ from $A$, $B$ and $C$ respectively. Prove that $\frac{kl}{h}$ is always the same regardless of the location of point $A$.

$h=\frac{S}{BC}$ and so on. Hence:
\[ \frac{kl}{h}=\frac{S\cdot BC}{AB \cdot AC}=\frac{1}{2}BC \sin{\theta} \]
Since $BC$ and $\theta$ are fixed, so the ratio is also fixed.

[i]Shobber[/i], you lost the number $2$ !

$bc=2Rh_a;\ ca=2Rh_b;\ ab=2Rh_c\Longrightarrow$
$\frac{h_bh_c}{h_a}=\frac{ac}{2R}\cdot \frac{ab}{2R}\cdot \frac{2R}{bc}=\frac{a^2}{2R}=2R\sin ^2A=R(1-\cos 2A).$

Yep, I forgot that $S=\frac{1}{2} \cdot h \cdot BC$...

Despite of it our answers are the same. $2R \cdot \sin{A}=BC$.  :)


geometry
3D geometry
tetrahedron
sphere
circumcircle

Hi, I need help with this problem.

Given a regular tetrahdreon that circumscribes four spheres, each of radius 8, find the point equidistant from all vertices.

[hide="Hint"]The point coincides with the circumcenter of the tetrahedron determined by the centers of the spheres, and that's easy to find by writing the equation based on Pythagorean Theorem.[/hide]


Suppose
\[ n(n\plus{}1)a_{n\plus{}1}\equal{}n(n\minus{}1)a_n\minus{}(n\minus{}2)a_{n\minus{}1}
\]
for every positive integer $ n\ge1$. Given that $ a_0\equal{}1,a_1\equal{}2$, find
\[ \frac{a_0}{a_1}\plus{}\frac{a_1}{a_2}\plus{}\frac{a_2}{a_3}\plus{}\dots\plus{}\frac{a_{50}}{a_{51}}.
\]

the solution which is seen above is true but the last term must be (n+1)
1/2+4+3+.......+(n+1)


induction
search
combinatorics unsolved
combinatorics

Prove that from any 200 integers it is possible to choose 100 numbers whose sum is divisible by 100.

In fact, from each set of 199 integers, you can choose 100 integers, the sum of which is divisible by 100. This is a special case of the following beautiful result: let $n$ be a positive integer. From a set of $2n - 1$ integers, one can always choose $n$ integers, the sum of which is divisible by $n$. (This has been discussed a few times on this forum.)

But how do we prove that from a set of any 2n-1 integers one can choose n integers whose sum is divisible by n?

By strong induction on $n$. First, try to prove it for the prime numbers.

(It's a tough problem!)

Try a search with keywords Erdös Ginzburg Ziv ...

Pierre.


[hide]10. Make a venn diagram or something.[/hide]

[quote="236factorial"]A class of 31 students was polled about which soft drink was their favorite. 17 students liked Fizz-Up, 18 liked Fizz-Down, and 6 students liked neither. How many students liked both?[/quote]

[hide="Click to reveal hidden content"]Since 6 people liked neither, we can just take them out of the total.

So 25 people liked Up and/or Down.

Because the given numbers add up to 35 (17+18), 10 people must like both (35-25)[/hide]

[quote="José"][quote="Wolfie"][hide]Since 6 liked neither, we can lower the student number to 25. Then, as 17+18=35, we do 35-25 and so ten students must have liked both.[/hide][/quote]

[hide] $\boxed {10}$[/hide][/quote]

You did [b]not[/b] need to do that; it said so explicitly in the solution.

[hide]10, all solutions were explained... :D [/hide]

[quote="236factorial"]A class of 31 students was polled about which soft drink was their favorite. 17 students liked Fizz-Up, 18 liked Fizz-Down, and 6 students liked neither. How many students liked both?[/quote][hide]

There is 31 in total.
First subtract the 6 students that didn't like any of them.
That is 25.
That means there are 25 students that like Fizz-up, Fizz-down, and both
Now add Fizz-up and Fizz-down.
That is 35. Do that because that is the total of people that like above, and both again.
Subtract 25 from 35.
That is 10.

10 is the answer.

[/hide]

[hide]17+18+6=41
To find the overlap, 41-31=10[/hide]


function
quadratics
inequalities
algebra unsolved
algebra

f is a trinom such that the equation f(x) = x has no real solutions .
show that the equation f(f(x)) = x has no real solutions as well.

What is a trinom?  If you just meant $ f: \mathbb{R}\to\mathbb{R}$ is some continuous function for which $ f(x) \equal{} x$ has no real solutions, then it's certainly true that $ f(f(x))\equal{}x$ has no real solutions, and the proof is as follows.

Since $ f$ is continuous, we have either $ f(x) > x$ for all $ x\in\mathbb{R}$ or $ f(x) < x$ for all $ x\in \mathbb{R}$.  Without loss of generality, assume $ f(x) > x$.  This implies that $ f(f(x)) > f(x) > x$ for all $ x$, and hence $ f(f(x)) \equal{} x$ has no real solutions.

thanks , a trinom is just a very particular case of continuous functions  :oops: :
f : x --> a x² + b x + c  trinom = polynom that has three terms .

Ah I see, we usually call it a quadratic where I'm from.  A trinomial is sometimes used to refer to a 3rd degree polynomial. :)

[quote="JoeBlow"]Since $ f$ is continuous, we have either $ f(x) > x$ for all $ x\in\mathbb{R}$ or $ f(x) < x$ for all $ x\in \mathbb{R}$.  [/quote]

Are you sure?

Of course that holds, if there are no solutions to $ f(x)\equal{}x$. You can just imagine it as a graph ---> either the whole function is above y=x or below it. Otherwise it'd have to intersect y=x somewhere, which it doesn't.

For a slightly different proof, assume that $ f(f(x_0)) \equal{} x_0$ for some $ x_0$, but $ f(x_0)\neq x_0$.  Then, assume $ f(x_0) > x_0$.  Therefore, calling $ y_0 \equal{} f(x_0)$, then $ f(y_0) \equal{} x_0 < y_0$.  Consider the obviously continuous function $ g(x) \equal{} f(x) \minus{} x$.  Obviously, $ g(x_0) > 0$, but $ g(y_0) < 0$.  By the intermediate value theorem, there is a $ c\in(x_0,y_0)$ such that $ g(c) \equal{} 0$, and $ f(c) \equal{} c$, contradiction.  Assume now  $ f(x_0) < x_0$ and repeat with inverted directions in all inequalities, done!  Note that we have not used that $ f(x)$ is a trinome, just that it is a continuous function!!!


sum of digits
algebra
Divisibility
modular arithmetic
arithmetic sequence
IMO Shortlist

Prove that for every real number $M$ there exists an infinite arithmetic progression such that:

- each term is a positive integer and the common difference is not divisible by 10
-  the sum of the digits of each term (in decimal representation) exceeds $M$.

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

WLOG let $ M$ be a positive integer (doesn't matter since we can just replace $ M$ with $ int(|M|) \plus{} 1$ and secure a result that is not weaker). Let the common difference be $ n \equal{} 10^k \minus{} 1$. Then I claim that for a sufficiently large value of $ k$, there is a starting value $ a$ such that no number with $ M$ or fewer non-zero digits can be congruent to $ a$ mod $ n$, showing that each value in the arithmetic progression $ a,a \plus{} n,a \plus{} 2n,...$ has at least $ M \plus{} 1$ non-zero digits, so has digital sum exceeding $ M$.

Each number with $ M$ or fewer non-zero digits can be written as $ \sum_{i \equal{} 1}^Ma_i\cdot 10^{b_i}$, with each $ a_i \in \{0,1,2,3,4,5,6,7,8,9\}$, and each $ b_i \in \mathbb{N}$. This expression can the take on at most $ 10^Mk^M$ different values mod n, since each $ a_i$ can take on at most 10 values, and each $ 10^{b_i}$ can only take on the values $ 10^0,10^1,...,10^{k \minus{} 1}$ ($ k$ different values) mod n. If $ 10^Mk^M < 10^k \minus{} 1$ then this guarantees an $ a$ so that each value in the arithmetic sequence has to be written using $ M \plus{} 1$ or more non-zero digits, securing the result. But this inequality will hold true for some large enough $ k$ since exponential functions grow faster than polynomial functions.

Seemed far too easy for an N6... maybe it's because it is 1999. More likely a mistake on my part, but I haven't found an error... please point out if there is one, I am very prone to making them.

Cheers,

Rofler

It obviously suffices to find arithmetic sequences for all positive integers $M$. Let $s(n)$ denote the sum of the digits of $n$ in base 10. We claim that for any positive integer $k$, $s((10^M - 1)k) \geq 9M$ (so the arithmetic sequence with [i]k[/i]th term $(10^M - 1)k$ would satisfy the conditions of the problem.) We will prove this by using strong induction on $k$.

For the base case $k = 1$, $s(10^M - 1) = 9M$. We now wish to show that for any $k > 1$, if our claim is true for all $j < k$, then $s((10^M - 1)k) \geq 9M$. If the base $10^M$ representation of $(10^M - 1)k$ is $a_0 10^0 + a_1 10^M + \cdots + a_p 10^{Mp}$, then $0 \equiv (10^M - 1)k = a_0 10^0 + a_1 10^M + \cdots + a_p 10^{Mp} \equiv a_0 + a_1 + \cdots + a_p \pmod{10^M - 1}$. Since $k > 1$, $p > 0$, so $a_0 + a_1 + \cdots + a_p < a_0 10^0 + a_1 10^M+ \cdots + a_p 10^{Mp}$, so we can find some $j < k$ such that $(10^M - 1)j = a_0 + a_1 + \cdots + a_p$. But
\begin{align*}
s((10^M - 1)k)
&= s(a_0 10^1 + a_1 10^M + \cdots + a_p 10^{Mp}) \\
&= s(a_0) + s(a_1) + \cdots + s(a_p) \\
&\geq s(a_0 + a_1 + \cdots + a_p) \\
&= s((10^M- 1)j) \\
&\geq 9M,
\end{align*}
which completes our proof.

Solution: We consider the number $c=\frac{10^k-1}{10-1}=111111111.....11$(with $k$ ones) where $k$ is a positive integer with $k>M$. Consider the AP $a_n=n*c$, where we put the said number we considered. We claim that $a_n$ has the desired property.
Lemma: For any $n$ if we try to write $a_n$ as sums of powers of $10$, at least $k$ powers of $10$ are used. This is equivalent to saying that sum of digits of $a_n$ are always at least $k$.
Proof: Suppose there exist nonnegative integers $p_1,p_2,p_3,...p_l$ such that $l \leq k-1$ and $c|\sum_{i=1}^{l} 10^{p_i}$. If so, we can suppose that all $p_i$ are less than or equal to $k-1$ since we know that $c|10^a-10^{a-k}$ trivially we can always replace $p_i$ with a smaller number if some $p_i$ is more than $k-1$.
This  trivially implies the existence of one number among $a_1, a_2, a_3,...a_9$ which has digit sum strictly less than $k$. This is a contradiction as trivially all $a_1, a_2,...a_9$(basically $111....1,222...2,....,999...9$ are the $a_1$ to $a_9$, each having $k$ digits) have digit sum at least $k$. So, we can see that all terms of the sequence have digit sum at least $k$. Proved.
Since $k>M$, this proves the problem.

Take any positive integer $M.$ Define a common difference $d = \overbrace{11 \dots 11}^{n \ 1s}$ for a very large $n.$ Consider any number that has sum of digits less than or equal to $M.$ Then it can be represented as the sum

$$\sum_{i=1}^M a_i$$

where each of $\{ a_1,a_2, \dots a_M \}$ is an integer power of $10,$ or $0.$ So each term may take on only $n+1$ different values $\pmod{d}$ since $10^n \equiv 10^{0} \pmod{d}.$ So there are at most $(n+1)^M$ different values this sum can take, but this is less than $d$ for large choices of $n.$ So there is some residue $a$ where any positive integer $\equiv a \pmod{d}$ has sum of digits exceeding $M$ as desired. $\square$

Simply take $a(10^k-1)$ for large $k$. It is easy to observe by writing $a_1a_2a_3…a_n000…0 - a_1a_2a_3…a_n$ that this has some of digits atleast $9k$. Use the fact $S(a) + S(b) \geq S(a+b)$ for this. And we are done!


algebra
polynomial
algorithm
Ring Theory
superior algebra
superior algebra unsolved

Define the ring $ R = \frac {\mathbb{Z}[X,Y]}{(XY - 1)}$. Show that $ (X^{*})$ is a maximal ideal of $ R$, where $ X^{*}$ is the class of $ X$ in $ R$. So $ (X^{*}) = X^{*} + (XY - 1)$.

I thought using that $ \frac {\mathbb{Z}[X,Y]}{(XY - 1)}$ is ismorphic with $ {\frac {(\mathbb{Z}[y])[X]}{(yX - 1)}}$ we get
$ (X^{*}) = \{f^{*}\ |f\ \in \mathbb{Z}[y]\}$. Is this correct ?

[quote="Amazigh"]I thought using that $ \frac {\mathbb{Z}[X,Y]}{(XY - 1)}$ is ismorphic with $ {\frac {(\mathbb{Z}[y])[X]}{(yX - 1)}}$ we get
$ (X^{*}) = \{f^{*}\ |f\ \in \mathbb{Z}[y]\}$. Is this correct ?[/quote]

It's correct, but I don't think that you've proved anything...

Actually, since $ X^*$ is a unit, we have $ (X^*)=R$, so I think the problem is wrong.

But $ (X^{*}) \equal{} \{f^{*}\ |f\ \in \mathbb{Z}[y]\}\equal{}R$, because every element in R we can write it as $ f^{*}\ |f\ \in \mathbb{Z}[y]$, if we use de dividing algoritm for communitative rings.

Sorry, I think I misread what you wrote.  Yes, I agree.  This ideal can't be maximal; it's the whole ring.

What is the division algorithm for commutative rings?  I wasn't aware that one existed.

But we don't need to use any division algorithm, just the definition of the ring in question.

I don't know of it calls "division algoritm", I mean that suppose we have polynomes f en g, then there exists q en r such that f=qg+r, with deg(r) ...

The question is wrong then I think. But how can u see that without using the above "algoritm" :)

Sorry, I keep missing your meaning.  OK, I'm not sure how the division algorithm proof goes (the problem I'm having is that $ yX\minus{}1$ does not have invertible leading coefficient) but here's the proof I had in mind:

Consider $ R/(X^*)$.  In this ring, we have $ \overline{X}\equal{}0$ and $ \overline{X}\cdot\overline{Y}\equal{}\overline{1}$, so $ \overline{1}\equal{}\overline{0}$ and $ R/(X^*)\equal{}0$.

Very nice.
By the way, it is not true that division algoritm, it would be true if we divided by a monic polynomial. The division algoritm is for a ring with a unity (1) AND dividing by a monic.

Small question, can you give an example of a prime ideal of R[x,y], it can't find one directly.

$ \mathbb{R}[x,y]$?  Or $ R$?  For the former, $ (x,y)$ will do, and for the latter, I believe the zero ideal is prime.

I am sorry, the question was to find an ideal which is NOT a prime :)
(in $ \mathbb{R}[x,y]$)

The square of any nontrivial ideal should do...


Ring Theory
superior algebra
superior algebra theorems

$F$ is a field. $F\langle x\rangle$ is the ring of all power series $f=a_0+a_1x+\cdots+a_nx^n+\cdots$ under the addition and multiplication defined as follows
$\sum a_ix^i + \sum b_ix^i = \sum(a_i+b_i)x^i$
and
$\sum a_ix^i \cdot \sum b_ix^i = \sum_{k=1}^\infty\left(\sum_{i+j=k} a_ib_j\right)x^i$

Show that $F\langle x\rangle$ has a unique maximal ideal.

The maximal ideal is the set of non-invertible elements. It's even principal.

Try to prove this, liyi:

A ring (assume it to be "nice": commutative, unitary, everything you want :)) is local (i.e. has a unique maximal ideal) iff the set of its non-invertible elements forms an ideal. In fact, for this problem, you only need the "if" part, which is also the easier part.  Now apply it here: the non-invertible series $a_0+a_1x+\ldots$ are precisely those which have $a_0=0$, so they obviously form an ideal.

As you have noticed, this ideal is also obviously generated by $x$.


The lengths in inches of the three sides of each of four triangles I, II, III, and IV are as follows:
\[ \begin{array}{rlrl} \hbox {I}& 3,\ 4,\ \hbox{and}\ 5 \qquad & \hbox{III}& 7,\ 24,\ \hbox{and}\ 25 \\ \hbox{II}& 4,\ 7\frac{1}{2},\ \hbox{and}\ 8\frac{1}{2} \qquad & \hbox{IV}& 3\frac{1}{2},\ 4\frac{1}{2},\ \hbox{and}\ 5\frac{1}{2}. \end{array}  \]
Of these four given triangles, the only right triangles are
\[ \begin{tabular}{rlrlrl} (A) & I and II \qquad & (B) & I and III \qquad & (C) & I and IV \\ (D) & I, II, and III \qquad & (E) & I, II, and IV & \end{tabular}  \]

[hide="About the 1972 Annual High School Contest"]
This is the contest, sponsored by MAA, that eventually became what we now know as the AMC contests.  At the time this was given, calculators were not allowed -- They barely even existed.

In 1972, there was no AMC-8, AMC-10, AMC-12, or AIME.  So, this contest contained a very wide range of questions.  Questions varied in point value from 3 points to 6 points.  The preceding question was from Part I (3-point questions).

Very good MATHCOUNTS students should be able to handle this problem.
[/hide]

[hide]The first one is 3,4,5. That is right.
The second is 8,15,17. Also right.
The third is 7,24,25. Right.
The fourth is 7,9,11. Not!!!
Hence, D.[/hide]

[hide]Use the Pythagorean theorem.
         3,4,5   Works
         4, 7.5, 8.5 Works
         7, 24, 25 Works
         3.5, 4.5, 5.5 NO
    Therefore, the answer is D.[/hide]

[hide]
First put them as integers first
$3,4,5$
$8,15,17$
$7,24,25$
$7,9,11$
All except the last are pythagorean triples.
$D$[/hide]

[hide]We know that I and III are definitely right triangles, as the lengths of their sides form a Pythagorean triple. For II, if we multiply all numbers by 2 to get whole numbers and it makes a triple, then we know II is a right triangle. After we do the same for IV, we see that only I, II, and III works, so answer D.[/hide]


Let $n$ be a natural number. Prove that neither $10^{n}$ nor $10^{n}+3$ can be written as a sum of the squares of three prime numbers.

$10^n \equiv 0 \mod4$ if $n>1$ but $p^2 \equiv 1 \mod 4$ when it is odd and $0$ for $2$ so ${q_1}^2+{q_2}^2+{q_3}^2 \equiv 3,2 \mod 4 $ so $10^n$t can't expressed as sum of squares of three prime nuber .
for $10$ is abvious .
for $10^n +3$ we consider both side mod 3.$10^n+3 \equiv 1 \mod 3$,
but ${q_1}^2+{q_2}^2+{q_3}^2 \equiv 0,2  \mod 3 $ and so no.


[quote="mindboggling"][quote]The handshake problem: I invite 10 couples to a party at my house. I ask everyone present, including my spouse, how many people they shook hands with. It turns out that everyone questioned (I didn’t question myself) shook hands with a different number of people. If we assume that no one shook hands with his or her spouse, how many people did my spouse shake hands with?

Please remember: I did not ask myself any questions.
Please also remember: Although it may seem so at first, this problem is NOT intractable.[/quote]

I kinda understand the question but can't think of any ways to solve this atm, any hints ?

Thanks.

Mon Oct 23, 2006 3:56 am
[hide="my futile attempt"] is the answer 9? First i listed the number of handshakes for ppl questioned 0 -> 18. What i found is that the person who shakes hand 18 times is coupled with the person who shakes 0 times, 17 with 1 and so forth. Only a person with 9 handshakes is left while the person who questions everyone is calculated to have made 9 handshakes, thus his wife made 9 handshakes.

However, I'm not 100% sure, still very confused[/hide][/quote]

im just wondering....did everyone only shake with everyone else once?? probably...but just wanna make sure. :maybe:


geometry
trapezoid
geometry proposed
U sviti matematyky

Let $ ABCD$ be a trapezoid. The circle $ w_1$ with center $ O_1$ is inscribed into the triangle $ ABD$, and the circle $ w_2$ with center $ O_2$ touches side $ CD$ and the extensions of sides $ BC$ and $ BD$ of triangle $ BCD$. It is known that $ AD\parallel{}O_1O_2\parallel{}BC$.

Prove that $ AC\equal{}O_1O_2$.

1.∠ADB=∠DBC  =>  ∠O1DB=∠O3BD  => D01∥B02
∠O1O2B=∠DB02   =>   D02B01 is an isoceles trapezium  =>
BD=O1O2

2. ∠O1DO2+∠O1BO2=180  but we have ∠O3DO2=90
  => ∠O1BO3+∠O3DO2=90    =>  ∠ABC+∠ADC=180    =>  ABCD is an isoceles trapezium
=> AC=BD
DONE!


here are some identities for which I would require not superior maths solutions! :D
thank you very much! :D

1) arctan(x)+arctan((1-x)/(1+x))=a
where a =-3pi/4 if x<-1 and a=pi/4 if x>-1.

2)arccosx=a
a=pi-arcsinsqrt(1-x^2) if x is in [-1,0)
and arcsinsqrt(1-x^2) if x is in [0,1].

thank you bye bye

1)

Solution:
Let u=arctanx and v=arctan((1-x)/(1+x)) so we get that tan(u+v)=1. If x<-1 then -pi/2 <u<-pi/4 and -pi/2<v<0 because x<-1 implies (1-x)/(1+x)<0, from which we get that -pi<u+v<-pi/4 which together with tan(u+v)=1, leads us to u+v=-3pi/4 so we get the first part.

If x>-1 then -pi/4<u<u/2 and -pi/2<v<pi/2 from which we get that -3pi/4<u+v<pi, which toghether with tan(u+v)=1, which gives that u+v=pi/4 so we get the 2nd part! :D :D

cheers! :D :D

2) Solution:

Lets denote u=arccosx. If x is in [-1,0) then cosu=x and pi/2<u<=pi which implies sinu=sqrt(1-x^2). because sin(pi-u)=sinu we get that sin(pi-u)=sqrt(1-x^2) and with o<=pi-u<pi/2 we get that pi-u=arcsin(sqrt(1-x^2)) so we get the 1st part.

you do the same for the 2nd part! :D

cheers! :D :D


R is asociative, not commutative ring, n>2 is fixed natural. if x^n=x for every x in R, prove that xy^(n-1)=y^(n-1)x for all x,y in R.


calculus
integration
function
quadratics
complex analysis

What is the value of
$ \int_{|z|\equal{}\rho} \frac {|dz|}{|z \minus{} a|^{2}}$,$ |a| \neq \rho$

No. It's the integral of a positive function over a curve of positive arclength.

We can transform this into a form more amenable to complex analysis. First, on the circle $ z\equal{}\rho e^{i\theta}$, $ |dz|\equal{}|i\rho e^{i\theta}|\,d\theta\equal{}\frac{dz}{i e^{i\theta}}\equal{}\frac{\rho}{iz}\,dz.$
Second, on that same circle, $ |z\minus{}a|^2 \equal{} (z\minus{}a)(\overline{z}\minus{}\overline{a}) \equal{} (z\minus{}a)\left(\frac{\rho^2}{z}\minus{}\overline{a}\right)\equal{}\minus{}\frac1z\left(\overline{a}z^2\minus{}(\rho^2\plus{}|a|^2)z\plus{}\rho^2 a\right).$

The full integral becomes $ \int_{|z|\equal{}\rho}\minus{}\frac{z}{(z\minus{}a)(\overline{a}z\minus{}\rho^2)}\cdot \frac{\rho}{iz}\,dz \equal{}\int_{|z|\equal{}\rho}\frac{i\rho}{(z\minus{}a)(\overline{a}z\minus{}\rho^2)}\,dz.$

The quadratic in the denominator always has one root inside the circle and one root outside, so the integral can be found by calculating the residue at exactly one simple pole.


probability
geometry
geometric transformation
reflection
logarithms
graph theory
probability and stats

Suppose that we have a graph with $ n$ vertices and no multiple edges. Start your random work at any vertex (i.e., choose the starting vertex in any way you want and at each moment go to one of the adjacent vertices to the vertex you are currently in with equal probability) and denote by $ T$ the first moment at which you visit some vertex for the second time. Prove that $ E[T]\le C\sqrt n$ with some absolute $ C$ (as usual, $ E$ stands for the expectation).

This I can solve and the solution is simple and short enough to make the question eligible for posting on AoPS :). So, let's see how much time it'll take the AoPS probability lovers to find it and whether somebody can come up with an approach different from the one I know ;).

What I cannot do now is to prove the same for the random walk with no backtracking, i.e., when you go with equal probability to each vertex except the one you just came from. Of course, for that case you have to assume something to exclude long linear pieces in the graph. The simplest assumption to add seems to be that the degree of each vertex is $ \ge 3$. I was pretty sure I had it at some point but when I tried to recall my argument, I couldn't do it :blush:.

The last remark is that the question seems very basic, so my feeling is that the answer should be well known for a long time. Still, Yuval Peres claimed that he couldn't find even the "easy" part in the literature, which puzzles me quite a bit... :?

Hi fedja

There's something that is not clear to me in the problem  :maybe: .

Let's say that we have k vertices in our graph, how much adjacent vertices do we have for each vertices ?

As you say that the probabilty to go to an adjacent vertice is uniform, it makes me think that we have a number of vertices which is something between 2 to k-1 for a given vertice. But is this number fixed or can it be variable for each vertices ?
For example if the number of adjacent is only two then we have a polygon. If the number is k-1 then we have the graph af all possible path between k point in the plane.

In any case it seems to be a complicated problem.

I am not sure but, as I understood it, Fedja was talking about a random walk on an arbitrary graph with $ n$ vertices (the graph is not regular). The bound $ C \sqrt {n}$ is tight since it is achieved with the complete graph with $ n$ vertices. Of course, in a random walks on the complete graph with $ n + 1$ vertices, $ P(T \geq k) = (1 - \frac {1}{n})(1 - \frac {2}{n}) \ldots (1 - \frac {k - 1}{n})$. But for $ 0 < k < \frac n2$, there exists a constant $ \lambda$ (does not depends on $ n$) such that $ 1 - \frac {k}{n} \geq \exp( - \lambda \frac {k}{n})$, hence $ P(T \geq k) \geq \exp( - \lambda \frac {k(k - 1)}{2n})$.
Now, because $ E(T) = \sum_k P(T\geq k)$ one can conclude that (for large $ n$)
$ E(t) > \sum_{k \leq \frac {1}{2}} \exp( - \lambda \frac {k(k - 1)}{2n}) \geq \sum_{k(k - 1) \leq n} \exp( - \lambda \frac {k(k - 1)}{2n}) \geq \sqrt {n} \exp( - \frac {\lambda}{2})$

Still have no idea how to prove the upper bound  :maybe:

Yes, the graph is completely arbitrary (except there should be at least one edge at each vertex, of course, plus, as I said, there should be no multiple edges). Once you are at a vertex $ v$ of degree $ d$, the probability to go to each adjacent vertex during the next step is $ 1/d$, exactly as alekk said.

[quote="fedja"]
This I can solve and the solution is simple and short enough to make the question eligible for posting on AoPS :). So, let's see how much time it'll take the AoPS probability lovers to find it and whether somebody can come up with an approach different from the one I know ;).
[/quote]
I've been thinking about this problem off and on for quite awhile, but I can't get a solution. You seem to have an easy solution which is a bit surprising to me... Are you inclined to post it? And if you have a solution for the backtracking random walk for degree $ > \sqrt{n}$ that should also work for the non-backtracking random walk because they're basically the same.
A professor tells me the result is known for families of expander graphs... http://en.wikipedia.org/wiki/Expander_graph. The proof uses the spectral definition.

I'm a bit too lazy to type the part that is needed for this particular case now, so here is the pdf file of the draft we wrote with Yuval (we consider a slightly more general problem there and there are a few stupid misprints in the text but, I believe, it is completely readable). If you have any questions, do not hesitate to ask :).

If you find a way to handle the non-backtracking case, I'll be most grateful. Nets Katz claimed that he had a counterexample about a month ago but has never sent me the details or confirmed his claim in any other way since then :(. Even if his counterexample was correct, it would change the $ \sqrt n$ bound just by a logarithmic factor.

Thank you for posting your solution. I've read it and more or less understood the details. But I'm also interested in how you came up with it... You defined  $ R_s \equal{}C\sqrt{k} \minus{}\frac{s}{\sqrt{k}} \minus{} \sum_z \psi(Q(z))$. Can you explain how come you started looking at this? and also how you get from that to (0.1)? I know that's a bit of a general question but I'm just trying to understand what your general idea was..

Of course, it all started with $ k\equal{}1$ (first self-crossing). Also, the very first case I looked at was the symmetric chain when $ p(x,y)\equal{}p(y,x)$. It was pretty obvious that in this case the expected time should depend on the set $ S_1$ (which at that stage was denoted just $ S$) and the last visited vertex $ x$. Now, it was fairly clear that the probabilities $ q(y)\equal{}\sum_{z\in S}p(y,z)$ (perhaps, together with the unvisited part of the graph structure, but this one was well beyond my ability to discern the right control for) were the main things to look at. So, the natural conjecture was that $ E[T]\le G(x,\{q(y)\})$ where $ G$ is some fancy function.

Disregarding the dependence of $ x$ for a moment, we can say that all other vertices should be treated equally, so $ G$ must be symmetric in $ q(y)$. The next thing was a big leap of faith: I assumed that $ G$ is additive in $ y$. This leads to the general form
\[ G\equal{}\sqrt n\left(C\minus{}\sum_{z}\psi(q(z))\right)\]
with some unknown $ \psi: [0,\plus{}\infty)\to[0,C/n]$.
(putting the normalization factor $ \sqrt n$ outside and writing $ C\minus{}\sum$ instead of $ \sum$ just reflect my personal habits of setting the formulae; there is nothing sacred about those).

This didn't work as written, so I was stuck for a while but then I recalled the idea used in writing the chess playing programs: once you have the rough estimate for the position, you can look one move ahead and use this rough estimate to evaluate the outcome after the first move. If you use this great idea, you get the factor $ 1\minus{}q(x)$ in front of your original $ G$ as the first correction immediately, and here you have a stroke of luck: this almost works (i.e., it works to the extent where it becomes more or less clear how to tune up the formula). Why we are lucky here, I don't really know.

At this stage, there was no logarithmic correction. The formula worked for the symmetric chains (with $ \psi(t)\equal{}\min\{t/\sqrt n, 1/n\}$ and also for almost symmetric chains. It didn't work only if $ p(x,y)$ differed from $ p(y,x)$ many times. Yuval suggested that we fix it by introducing that logarithm. It was a rather natural suggestion in the hindsight, but if you want to know how he came up with it, you should ask him, not me :P. The old $ \psi$ didn't work well with this correction, so we had to create a new one (which was done by just figuring out what exactly should be the properties it must satisfy; this was more or less routine).

Now we had the estimate for the first self-intersection and the only thing that remained was to set up a reasonable general expression that reduces to this estimate for $ k\equal{}1$. That is a story of trial and error but really we had just a finite choice problem here. The expression in that draft is just the cleanest and most structured working formula we could come up with but its exact shape is "historically conditioned" (i.e., if it were you instead of us, it would, probably, look slightly different :)).

This story is a bit "idealized" in the sense that it shows how I would think it up if I were really, really clever (the reality is that I'm not, so there had been a few unmentioned dead-end attempts that failed but finally pushed me to "the right track").

Thank you for that really good explanation. I was confused because of this general solution for any $ k$ but of course I should have just tried to understand it for $ k\equal{}1$.
I'm also trying to adapt your proof to the non-backtracking case. I'm thinking of looking at new transition probabilities. Instead of taking one step, take some constant number of steps $ K$. ie, $ \tilde {p}(x,y) \equal{} p^K(x,y)$. But I have a hunch you already looked at this since you mentioned that there was a possibility of a counterexample but that you knew that the bound had at most a $ \log n$ factor added. And I think this follows from taking $ K \equal{} \log n$. Should I be looking for another method?

It is unclear. Taking $ K$ just a big constant doesn't help much because the graph can have the tree structure on that scale. Taking $ K\gg \log n$ will, probably, work but I haven't really tried that because I hoped for a pure $ \sqrt n$. So, go ahead and try this way. Let me know if you come up with something interesting :).


LaTeX
combinatorics unsolved
combinatorics

Let $M= \{ 1, 2, \cdots, 19 \}$ and $A = \{ a_{1}, a_{2}, \cdots, a_{k}\}\subseteq M$. Find the least $k$ so that for any $b \in M$, there exist $a_{i}, a_{j}\in A$, satisfying $b=a_{i}$ or $b=a_{i}\pm a_{i}$ ($a_{i}$ and $a_{j}$ do not have to be different) .

[quote=ychjae]Please check for my idea :| :
If $k\leq 3$, then there are at most three elements in $A$, let them be $a<b<c$ (because I am too lasy to type $a_{1},a_{2},a_{3}...$ in LaTex :P ), and $A$ can only "produce" not more than $12$ integer range between $1$ and $19$, i.e. $a,b,c,2a,2b,2c,a+b,a+c,b+c,c-a,c-b,b-a$, so $k\geq 4$.
  For $k=4$,let the $4$ elements be $a<b<c<d$, then there are at most $20$ integers then range between $1$ and $19$ which is "produce" from $A$,i.e.
$X=\{a,b,c,d,2a,2b,2c,2d,a+b,a+c,a+d,b+c,b+d,c+d,d-a,d-b,d-c,c-b,c-a,b-a\}$
so there are at most one pair of overlap elements xor one out-of-range element in $X$.
  As $2d$ should not less than $19$(or else no other combination is big enough to produce $19$) but $2d\neq 19$,we know no two element in $X$ can overlap and $d+c=19$.
  On the other hand,$c\geq 3$ so $10\leq d \leq16$ and $c=19-d$.
 Case 1:$d=10,c=9$,we have $b=7$ ,try all possibe $a$,no solution.
 Case 2:$d=11,12,13,14,15,16$ and $c=19-d$,only $d+b$ can produce $18$,so $b=c-1$, try all possible $a$ ,no solution.
  For $k=5$,$A=\{2,6,7,8,11\}$  works.
P.S.1 My solution to prove $k\neq 4$ is extremely ugly,hope someone can improve it.
P.S.2 Consider all possibe parity combination for $k=4$ and with the fact that there are 10 odd number in $M$,we can show there should be 2 odd and 2 even in $A$ so this can reduce the times of try and error.[/quote]

After getting $d+c=19$ you can directly say $c=9$ because second biggest element of $X$ must be $2c$ or $b+d$ but $17\neq 2c$ so $2c=18 \implies c=9$. After that we can easily get $b=7$ and $a=4$ which gives a contradiction ($d+a=2b$).
I did the same for the other parts of the problem. I honestly don't think there is an easier way.

Another example for $k=5$:
$A=\{1,2,4,10,17\}$


geometry
geometric transformation
reflection
rectangle
circumcircle
geometry solved

Let $ABCD$ is a quadrilateral with $AC \perp BD$. AC cut BD at S. M,N,P,Q is reflection with S through AB,BC,CD,DA. (SPQ) cut AP at E. (SMN) cut AN at F. Prove that MEFQ is an inscribed quadrilateral.

You can find solution in the book "Plane Geometry" written by Prasolov. Problem nr 2.77 in english version.

2.77
"Let P be a intersection of diagonals of ABCD. Prove that the midpoints of the sides of quadrilateral ABCD and the projections of point P on the sides lie on one circle."
Solution
2.77. The midpoints of the sides of quadrilateral ABCD are vertices of a rectangle
hence, they lie on one circle. Let K and L be the midpoints
of sides AB and CD, let M be the intersection point of lines KP and CD. By
Problem 2.76 PM is perpendicular to CD; hence, M is the projection of point P on side CD and
point M lies on the circle with diameter KL.
For the other projections the proof is similar.

You can find Prasolov book for example on http://matex2004a.staszic.waw.pl/prasolov-%20planimetry%20pt%201.pdf

I think your problem is different from mine!!!  :!:

No, it's the same. In my problem we prove that 4 more point lie on this circle.

I am not convinced either that the problem posted here is equivalent to Prasolov's problem 2.77 (which is actually the well-known [url=http://www.cut-the-knot.org/Curriculum/Geometry/EightPointCircle.shtml]eight point circle theorem[/url]). However, the problem is not particularly hard to solve:

[color=blue][b]Problem.[/b] Let ABCD be a quadrilateral whose diagonals AC and BD are perpendicular to each other and intersect each other at a point S. Let M, N, P, Q be the reflections of the point S in the lines AB, BC, CD, DA, respectively. The circumcircle of triangle SPQ intersects the line AP at a point E (apart from P), and the circumcircle of triangle SMN intersects the line AN at a point F (apart from N). Show that the points M, F, E, Q lie on one circle.[/color]

[i]Solution.[/i] Since the point M is the reflection of the point S in the line AB, we have BM = BS. Similarly, BN = BS. Thus, BM = BN = BS, so that the point B is the center of the circumcircle of triangle SMN. Since the point S lies on this circumcircle and $AS\perp BS$ (this is equivalent to $AC\perp BD$), the line AS must be the tangent to this circumcircle at the point S, and thus, by the intersecting chord and tangent theorem, $AN\cdot AF=AS^2$. Thus, the point F is the image of the point N under the inversion with respect to the circle with center A and radius AS. Similarly, the point E is the image of the point P under this inversion. Also, since the point M is the reflection of the point S in the line AB, we have AM = AS, so that the point M lies on the circle with center A and radius AS, and thus this point M is its own image under the inversion with respect to this circle; for similar reasons, the point Q is its own image under this inversion. Altogether, the inversion with respect to the circle with center A and radius AS maps the points M, N, P, Q to the points M, F, E, Q, respectively. In http://www.mathlinks.ro/Forum/viewtopic.php?t=4344 , it was shown that the points M, N, P, Q lie on one circle; thus, their inversive images, i. e. the points M, F, E, Q, must also lie on one circle, and we are done.

What you also get, by the way, is that the points N, P, F, E lie on one circle; this follows from the converse of the intersecting chords theorem, since $AN\cdot AF=AS^2$ and the similarly obtained equation $AP\cdot AE=AS^2$ yield $AN\cdot AF=AP\cdot AE$.

  Darij


integration
logarithms
calculus
real analysis
real analysis unsolved

The integral equals the famous Catalan constant $ C$.

You may want to have a look at this collection of formulas:

http://www.cs.cmu.edu/~adamchik/articles/catalan/catalan.htm

especially the integral formula after Entry 33.


trigonometry
geometry
perpendicular bisector
angle bisector
Pythagorean Theorem
trig identities
Law of Sines

In triangle ABC, $ \angle ABC\equal{}75^{\circ}$, $ \angle BAC\equal{}60^{\circ}$, and $ AB\equal{}8$. Let D be the intersection point of BC's perpendicular bisector and BAC's angle bisector.

Find the value of DC.

Here's my solution: [hide]Draw an altitude from point B to side AC. Let F be the point where the altitude intersects side AC. Note that F must also be the point where the perpendicular bisector of BC intersects AC. Then triangle ABF is a $ 30^{\circ}\minus{}60^{\circ}\minus{}90^{\circ}$ right triangle and BF is $ 4\sqrt{3}$. Since angle FBC is $ 45^{\circ}$ and angle BFC is right, angle BCF must be $ 45^{\circ}$ also. Thus, triangle BFC is a $ 45^{\circ}\minus{}45^{\circ}\minus{}90^{\circ}$ right triangle and FC is also $ 4\sqrt{3}$. Let H be the point where FD intersects BC. Clearly, triangle FHC is also a $ 45^{\circ}\minus{}45^{\circ}\minus{}90^{\circ}$ right triangle. Thus, $ HC \equal{} 2\sqrt{6}$.

Let G be the point where the angle bisector of BAC intersects side BC. Since angle BAG is $ 30^{\circ}$ and angle ABC is $ 75^{\circ}$, angle BGA must be $ 75^{\circ}$. Thus, triangle ABG is an isosceles triangle and AG must also be 8.

Now, consider triangle AGC. Using the law of sines, we can find the length of side GC as follows:

$ \frac{GC}{\sin 30^{\circ}} \equal{} \frac{8}{\sin 45^{\circ}}$

This gives us $ GC \equal{} 4\sqrt{2}$.

Now, consider triangle DGC. Since $ GC \equal{} 4\sqrt{2}$ and $ HC \equal{} 2\sqrt{6}$, $ GH \equal{} 4\sqrt{2} \minus{} 2\sqrt{6}$. [ Note that angle DGH and angle BGA are vertical angles, so angle DGH is also $ 75^{\circ}$. Then, angle GDH must be the complement, $ 15^{\circ}$. ] In right triangle GDH:

$ \tan 15^{\circ} \equal{} \frac{GH}{DH} \equal{} \frac{4\sqrt{2} \minus{} 2\sqrt{6}}{DH}$

Noting that $ \tan 15^{\circ} \equal{} 2 \minus{} \sqrt{3}$, and solving for DH, we get $ DH \equal{} 2\sqrt{2}$. Then applying the Pythagorean theorem to triangle DHC, we get $ DC \equal{} 4\sqrt{2}$.

I note that triangle DGC is isosceles so I'm sure there must be an easier way to solve this problem. I think I must be missing something simple here.

I wonder if reflecting point A over line DF might be a helpful strategy. Also, might an easier solution involve finding BD instead of DC (since they are congruent line segments)?
[/hide]

Did anyone find a better way?

[hide="May or may not be helpful"]It just so happens that $ ABDC$ is cyclic...[/hide]

[hide="easier solution"]
So, label the intersection of the perpendicular bisector of $ BC$ and $ BC$ as $ E$. Label the intersection of $ DE$ and $ AC$ as $ F$. Then, draw $ BF$. Obviously, $ BFE$ is a 45-45-90 triangle, and $ BAF$ is a 60-30-90. Since $ AB$ is 8, then $ AF \equal{} 4$ and $ BF \equal{} 4\sqrt{3}$. Consequently, since $ BFE$ is 45-45-90, $ BE \equal{} 2\sqrt{6}$. $ BD \equal{} DC$ because $ B$ and $ C$ are equidistant from the perpendicular bisector. Then, use ptolemy's theorem!

You get \[ x(4)(1\plus{}\sqrt{3}) \plus{} 8x \equal{} 4(1\plus{}\sqrt{3})(4\sqrt{6}).\]

That simplifies to $ x\equal{}4\sqrt{2}$.
[/hide]

Let $ \angle BCD \equal{} \angle CBD \equal{} x$. Using the lemma
$ \frac {sin BAD}{sin DAC} \cdot \frac {sin ACB}{sin BCD} \cdot \frac {sin CDA}{sin ADB} \cdot \frac {sin DBC}{sin CBA} \equal{} 1$, we have
$ sin45sin(105 \minus{} x) \equal{} sin75sin(75 \minus{} x)$
From here we get $ x \equal{} 30$, as grn_turtle's statement implies. So by the law of sines on $ \triangle ABC$, $ BC \equal{} 4\sqrt {6}$, from which we easily get $ CD \equal{} 4\sqrt {2}$.

By the way, for red p_nguin's solution, Ptolemy is unneccesary, you know BDE and CED are congruent 30-60-90 triangles.


geometry
geometric transformation
reflection
search
probability
algorithm
algebra

Well does anyone know faster nqueen algo? (Those who practice in USACO they have faced checker’s problem.)

Anyhow it is an obvious optimization of halving the searching area by using the reflection idea. But I think there is more something to do to make the code much faster. Does anyone know that?

I believe mahbub is referring to the problem of placing n queens on a n-by-n chessboard such that none of the queens attack another.  The nqueens problem is the problem of finding all such arrangement of queens given an n-by-n board, if I remember correctly.

i used a depth first search with boolean arrays representing the board.  are you thinking of doing it faster than that?

Yes i want much faster algo. Bcoz that is simple idea to use backtraking (it is same as DFS) with flaggingand with the idea of reflection to half the searching space. But i know it can be reduced more. But i don't know how. So if anyone know plz tell here!

i seen the problem solved for 100 queens on a normal computer in less than 1 second.
let's see if i remember the idea.
So, first you could try a simple one. just arrange the dames random. so you put first queen random on first row, then second queen random on the available positions of second row and so on. if you get stuck you just start all over again from the first queen. (the idea is about the probability of getting a correct answer)
if it doesn't work you can try this:
you just put the first queen random, then you put the second queen anywhere on the second row but such that the following queens have a maximum number of possibilities left (there could be more than one square on second row such that following  queen has a maximum number of possibilities; you choose one random if this is the case), then you put the third queen on the third row such that the fourth queen has a maximum number of possibilities and so on. If you get stuck, you start all over again from the first queen. By "the following queens have a maximum number of possibilities left" i mean that the number of cells attacked by the new queen and which are not attacked by the previous queens is minimum.
And you can repeat what i said above this time leaving a minimum number of possibilities for the next queen.
I know the algorithms are not deterministic, but as i know are the best in practice, and they are also used (the second and third) for the horse problem (start with a horse from any square of a chess table and try to cover all the squares without going twice in the same square).

I think the difficulty of this problem comes from the fact that you have to find all of the placements, not just one.
Though finding an arrangement for 100 still seems like a daunting task...

trying to find all combinations is an NP-complete problem. It means that there is no polynomail algorithm know to solve the problem. Actually trying to find just one solution with a deterministic algorithm is an NP-complete problem also.
If any of you finds a polynomial solution he is close to getting himself a Nobel prize. :). So, good luck!

hey, i did not ask for a polynomial algo. i asked for a faster algo which use the idea of symmetry. if u go to usaco u will see a problem asking for all the arrangement. there many solved the problem in vey short time. i was wondering of this. so i asked here.

i searched for just single soln for nqueen problem for n about 2 weeks ago. i saw a O(1) soln. just giving u pattern. :rotfl: so if u wanna to get a single soln of a particular case u don't need to do such work! :D

So i am still looking for a faster algo for this prob. actually they said use reflection and rotation. well, reflection is ez one. but what about roattion? remember we are asked to find all the soln!  :(

all i wanted to say is that even if you use rotation or whatever techinque, if you want a deterministic algorithm it will still be exponential.
if you want a much faster solution than others, you can use the random one i told you about or maybe try to find some other random algorithms.

can anybody tell me what soln is and a site for this thing? :)

oh u can't see that soln at that site. But well i am sending PM to you. check that.

thanks! i'll take a look on it. till then, it says there somewhere that the program given by that ucrainean guy works 4 times faster than ussual techinques. But now since any known algorithm has a factorial time of solving you realise, 4 times faster means practically nothing since for example switching from N=13 queens to N=14 queens will increase the time 14 times. So, ok, i admit it's a nice gain, but still nothing real.
As i heard there is a romanian teacher called Paun working in Spain (Salamanca i guess but i am not sure, maybe i am wrong), and together with a team he was trying to build somekind of a biological computer, :), to solve the NP-complete problems in polynomail time somehow using factorial memory(so when N is increasing, instead of increasing the time it took the algorithm to solve the problem N times, he wanted to increase the memeory the algorithm was using N times). It was a nice idea, and if you are interested i can tell you more about it (anyway i don't know too much myself), still, as i understood at a confference he held in romania, they didn't succeed to good. Still, the idea was nice. :)

well, if u r quite familiar with contest programming then u will understand that this 4time faster code is a very good achievement. :D  it is not something to make general faster algo for all n. it is just imporving searching time. NOTE: searching time. that means i will do brute force work but i want to do in such a way that it will take less time. :P

i am familliar with programming contests, but i never thought that they still propose NP comlpete problems in there. :)

yeh, it is very common to give NP problem in programming contest :lol:

But almost every time the judges are quite sympathetic to the contestants so they do not look for the best possible algo. But it is always good to learn the best one! ;)

Hate optimisation tasks... They are just frustating. :P

Lucky that IOI is mainly a algo-competition. ;)

Such a hard problem needs a little bit of artificial intelligence; with heuristic functions and a very high probability ( hoping that it finds the solution), you can get a much faster result. I used to program a simpler game using AlphaBeta pruning technique and it worked very nicely. But this is not suitable for N-Queens problem.

[quote="mahbub"]hey, there is a good news. usaco published 3 new solns of this problem. the first one uses reflection idea which is purely official soln. And the rest two r from two users one from romania and other i don't know. i haven't checked those yet. i will read that later. thanks to you every1 for posting many posting in this topic! :)[/quote]

Excuse me but can I get those solns?
I'm very interested in it.

In addition to this, I really wonder if threre exists the proof of the N-Queen problem,
supporting that for all N >= 4, there exists solution for it.

thank you :)

although i don't know the proof i am sure it exists, moreover the number of solutions for a given n i think it's factorial or so. i'll think maybe i can compe up with a proof myself. shouldn't be too hard.

i think proving the existence of a single unattacking board is ez. but o find all the soln is tough and i am thinking impossible! :P  But if u can then i am really very interested to have that! So well, do u need the proof of the exitence of single board?

Well, there's a mathematical answer to that task (not allowed in USACO though!)

You can read about it here: http://mathworld.wolfram.com/QueensProblem.html


induction
combinatorics proposed
combinatorics

Prove that for arbitary positive integer $ n\geq 4$, there exists a permutation of the subsets that contain at least two elements of the set $ G_{n} \equal{} \{1,2,3,\cdots,n\}$: $ P_{1},P_{2},\cdots,P_{2^n \minus{} n \minus{} 1}$ such that $ |P_{i}\cap P_{i \plus{} 1}| \equal{} 2,i \equal{} 1,2,\cdots,2^n \minus{} n \minus{} 2.$

We call the permutation of subsets of set $ G_n$ described in problem a type 2 permutation.
Lemma: There is a permutation of all (except empty set) subsets of the set $ G_n \equal{} \{1,2,...,n\}$: $ P_1,P_2,...,P_{2^n \minus{} 1}$ such that $ P_1 \equal{} \{1,2,...,n \minus{} 1\}$, $ P_{2^n \minus{} 1} \equal{} n$, $ P_{2^n \minus{} 2} \not \equal{} G_n$ and $ |P_i \cap P_{i \plus{} 1}| \equal{} 1$. - type 1 permutation

If we prove this lemma, then by induction, we can solve the problem:
We prove by induction that there is type 2 permutation of $ G_n$ but with extra conditions, that starts and ends with two-element subsets, and the last subset doesn't contain $ n$. Suppose that there exist such permutation for $ G_n$ and let's prove it exists for $ G_{n \plus{} 1}$. By lemma there is type 1 permutation of $ G_n$. Add $ n \plus{} 1$ to every set in that permutation, then first subset is $ \{1,2,...,n \minus{} 1,n \plus{} 1\}$ and the last one is $ \{n,n \plus{} 1\}$ and every two consecutive subsets have two elements in common now (the one they had before and $ n \plus{} 1$) - in this way we got some sequence of subsets of $ G_{n \plus{} 1}$, call it $ Q$. By induction, there is type 2 permutation of $ G_n$ in which last element is two-element subset which doesn't contain $ n$, and let this permutation be $ R$. Those subsets in $ R$ and $ Q$ are in fact all subsets of $ G_{n \plus{} 1}$ that have at least two elements. Then just write permutation $ R$ and after it permutation $ Q$. Last element in $ R$ is two-element subset which doesn't contain $ n$, so it has two elements in common with first subset in $ Q$ which is $ \{1,2,...,n \minus{} 1,n \plus{} 1\}$. But this permutation $ RQ$ ends with set $ \{n,n \plus{} 1\}$ and we don't want that, so just look at permutation written in inverse order, and that's it, we got desirable permutation for $ G_{n \plus{} 1}$.

Now, let's prove this lemma. This prove also goes by induction. Observe type 1 permutation of $ G_n$, $ P_1, P_2,..., P_{2^n \minus{} 1}$ where $ P_1 \equal{} \{1,2,...,n \minus{} 1\}$ and $ P_{2^n \minus{} 1} \equal{} \{n\}$ but $ P_{2^n \minus{} 2}\not \equal{} G_n$. Write it twice, but second time in inverse order $ P_1, P_2, P_3, ... , P_{2^n \minus{} 1}, P_{2^n \minus{} 1}, P_{2^n \minus{} 2}, ... ,P_1$. In this sequnce again any two consecutive sets have one element in common. Now, start from the begginig and add $ n \plus{} 1$ to second, forth, sixth, ..., $ 2^{n \plus{} 1} \minus{} 2$th subset, and at the end write $ n \plus{} 1$. Observe that we didn't add $ n \plus{} 1$ to any consecutive subsets, but we got all subsets of $ G_{n \plus{} 1}$.
What's more is that we get last subset to be $ \{n \plus{} 1\}$, the subset before him isn't $ G_{n \plus{} 1}$ (it's $ \{1,2,...,n \minus{} 1,n \plus{} 1\}$) and everything we need is to get the first subset to be $ \{1,2,...,n\}$. First subset in this sequence is $ \{1,2,...,n \minus{} 1\}$, and the subset $ \{1,2,...,n\}$ by induction wasn't adjecent with subset that contains $ n$ before adding $ n \plus{} 1$ to all subsets. So because we didn't add $ n \plus{} 1$ in first subset, we can just swich the first and one of the $ \{1,2,...,n\}$ in which we didn't add $ n \plus{} 1$, and that's it.

My English is poor so just ask if you don't get something, or I made some mistakes

Bye

Example: $ n \equal{} 3$, type 1: $ 12, 1, 123, 2, 23, 13, 3$, write it twice and add $ 4$ to those on even places $ 12, 1(4), 123, 2(4), 23, 13(4), 3, 3(4), 13, 23(4), 2, 123(4), 1, 12(4), (4)$, now swich $ 12$ and $ 123$.
type 2 for $ n \equal{} 4$: $ 14, 124, 12, 123, 13, 134, 34, 234, 24, 1234, 23$.
For $ n \equal{} 5$ it's $ 14, 124, 12, 123, 13, 134, 34, 234, 24, 1234, 23$ + $ 1235, 1(4)5, 125, 2(4)5, 235, 13(4)5, 35, 3(4)5, 135, 23(4)5, 25, 123(4)5, 15, 12(4)5, (4)5$
and just write it in inverse order :)

[quote="SpongeBob"] By induction, there is type 2 permutation of $ G_n$ in which last element is two-element subset which doesn't contain $ n$, and let this permutation be $ R$. Those subsets in $ R$ and $ Q$ are in fact all subsets of $ G_{n \plus{} 1}$ that have at least two elements. Then just write permutation $ R$ and after it permutation $ Q$. Last element in $ R$ is two-element subset which doesn't contain $ n$, so it has two elements in common with first subset in $ Q$ which is $ \{1,2,...,n \minus{} 1,n \plus{} 1\}$. But this permutation $ RQ$ ends with set $ \{n,n \plus{} 1\}$ and we don't want that, so just look at permutation written in inverse order, and that's it, we got desirable permutation for $ G_{n \plus{} 1}$.

[/quote]
 ends with set $ \{n,n \plus{} 1\}$ and we don't want that, so just look at permutation written in inverse order, and that's it, we got desirable permutation for $ G_{n \plus{} 1}$.
why cant the set end with set $ \{n,n \plus{} 1\}$


Find all the positive integers n , to make (2^n+1)/(n^2) an integer.
thanks,i need it very soon.

but how to prove,where posted?
can you tell me ,thanks very very very very much

IMO 1990 #3: http://www.mathlinks.ro/viewtopic.php?t=60740

PEN A71: http://www.mathlinks.ro/viewtopic.php?t=150439


$ 1)$aadade haghighi va namanfie $ a,b,c$ mafroozand be torike $ a \plus{} b \plus{} c \equal{} 2$ sabet konid:
$ (a^2 \plus{} ab \plus{} b^2)(b^2 \plus{} bc \plus{} c^2)(c^2 \plus{} ca \plus{} a^2)\leq 3$

$ 2)$dar mosalase $ \triangle ABC$ farz konid $ BC \equal{} a,CA \equal{} b,AB \equal{} c$ sabet konid:
$ \frac {3(a^4 \plus{} b^4 \plus{} c^4)}{(a^2 \plus{} b^2 \plus{} c^2)^2} \plus{} \frac {ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2}\geq 2$

$ 3)$aadade sahih va mosbate $ a,b,c$ mafroozand be torike $ a \plus{} b \plus{} c \equal{} 3$ sabet konid:
$ \frac {1 \plus{} b \plus{} c^2}{1 \plus{} c \plus{} b^2} \plus{} \frac {1 \plus{} c \plus{} a^2}{1 \plus{} a \plus{} c^2} \plus{} \frac {1 \plus{} a \plus{} b^2}{1 \plus{} b \plus{} a^2}\geq 3$

khob..post poste namosavie,manam goftam 1 soal bedam.soale bahal o jalebie:
a,b,c>0  sabet konid sigma dorie a^2.b^2 / a+35b <= a^3+b^3+c^3 / 18

rastesh in ye marathone namosavi nis ke toosh soalaye na,mosavi bezarim,man in topico doros kardam ke roo in 3ta bahs konim,age too ye topice mojaza roo soale shoma bahs konim fek konam behtar bashe  :)

[quote="khashi70"]fekr konam dovvomi baraye har 3 ta adade mosbat doroste , yani lazem nist azlae mosalas bashan , intor nist ?[/quote]
are baraye tamame aadade mosbat doroste...
vali ba estefade az taghir moteghaiere $ a \equal{} x \plus{} y,b \equal{} y \plus{} z,c \equal{} z \plus{} x$ ham mishe sabetesh kardo ehtemalan manzoore tarah ham hamin boode...

babak namosavie $ 3$ fekr konam eshkale typi dare !

manzooret [b]sahih[/b]eshe?
oono khodetoon bokonidesh haghighi,are injash eshteba shode...

man $ 1$ o hal kardam vali raham kheyli kasife , rahe ghashang dare ?

bastegi dare rahe ghashang be chi begi... :)
to ba muirhead va schur hallesh kardi are? :maybe:


inequalities
calculus
integration
trigonometry
inequalities solved

I must be blind because I really don't see... in spite of the indication :( :what?: :mad:.

Jordan's Lemma?  :P

$e^{-R\sin x}\leq e^{-\frac {2R}{\pi}x}$

Integrating, we get:

$\frac {\pi}{2R}\cdot (1-e^{-R})$

You can use this to show that given a contour in the upper-half plane and given that $a>0$, the portion on the upper semi-circle ($\int_C f(z)e^{iaz}$) goes to $0$ as $R\to\infty$ as long as $f(z)\to 0$ as $R\to\infty$


Bob fired a rocket from his home. He then leaves his home and drives 40 east and then 50 south and
then 10 west and finds the rocket. What is the approximate distance the rocket traveled from his home?

A) 30 mi B) 40 mi C) 58 mi D) 64 mi E) 80 mi

Bob travels a net distance of $ 30$ mi east and $ 50$ mi south. The distance from his house by the Pythagorean Theorem is then $ \sqrt{30^2\plus{}50^2}\equal{}\sqrt{3400}\approx58$. So the answer is C.


parameterization
function
limit
logarithms
calculus
calculus computations

[b]1. The problem statement, all variables and given/known data[/b]

Find all values of the parameter a>0 such that the function

$ f(x)=\{\begin{array}{cc}\frac{a^x+a^{-x}-2}{x^2},x>0\\3ln(a-x)-2,x\leq0\end{array}$

[b]3. The attempt at a solution[/b]

$ \lim_{x\rightarrow 0}\frac{a^x+a^{-x}-2}{x^2}=0$

$ 0=3ln(a-0)-2\rightarrow a=e^{2/3}$

[quote="azatkgz"][b]1. The problem statement, all variables and given/known data[/b]

Find all values of the parameter a>0 such that the function

$ f(x) = \{\begin{array}{cc}\frac {a^x + a^{ - x} - 2}{x^2},x > 0 \\
3ln(a - x) - 2,x\leq0\end{array}$

[b]3. The attempt at a solution[/b]

$ \lim_{x\rightarrow 0}\frac {a^x + a^{ - x} - 2}{x^2} = 0$

$ 0 = 3ln(a - 0) - 2\rightarrow a = e^{2/3}$[/quote]
this is wrong
$ \lim_{x\rightarrow 0}\frac {a^x + a^{ - x} - 2}{x^2} = 0$

the limit is equal to $ \ln ^{2} a$

[quote="azatkgz"]I'll try solve.[/quote]
if you know Taylor expansion then it's obviusly. if you don't know and know Hospital rule then use this rule one times here
 $ \lim _{x \minus{} > 0} \frac {a^{x} \minus{} 1}{x} \equal{} \ln a$ ( it's also  well known )
and here $ \lim _{x \minus{} > 0} \frac {\frac {a^{x} \minus{} 1}{x} \minus{} \ln a }{x} \equal{} \frac {\ln ^{2}a}{2}$
if you don't know Hospital rule, then you must study this rule, you can try to find in some math analysis book


Any hint for the following problem? [Don't give me the full solution; I want to solve it myself!]

The function $ f(x)$ satisfies $ f(2\plus{}x)\equal{}f(2\minus{}x)$ for all real numbers $ x$. If the equation $ f(x)\equal{}0$ has exactly 4 distinct real roots, find the sum of the roots?

Using Vieta?

Well, write the condition as $ f(x)\equal{}f(4\minus{}x)$

Now, if $ f(k)\equal{}0$ then we must also have $ f(4\minus{}k)\equal{}0$

Now, if k=2, we have an odd number of roots by pairing, so that just cannot happen.

So, what would happen otherwise... lol.

[quote="ThetaPi"]Any hint for the following problem? [Don't give me the full solution; I want to solve it myself!]

The function $ f(x)$ satisfies $ f(2 \plus{} x) \equal{} f(2 \minus{} x)$ for all real numbers $ x$. If the equation $ f(x) \equal{} 0$ has exactly 4 distinct real roots, find the sum of the roots?

Using Vieta?[/quote]
Hint [hide]Your function is obviously symmetrical around $ x\equal{}2$.
Whatever root is to one side, there is another one to the other side. If it had an odd number of roots, $ x\equal{}2$ would be one of them.
In other words, as rofler put it, if $ k$ is a root, so is $ 4\minus{}k$.[/hide]
Further hint [hide]If $ k$ and $ 4\minus{}k$ are two of the roots, there must be another pair of roots: $ m$ and $ 4\minus{}m$.[/hide]
Alternate further hint and more: [hide]If $ f(x)$ is symmetrical around $ x\equal{}a$
$ ($ or the function $ f(x)$ satisfies $ f(a \plus{} x) \equal{} f(a \minus{} x)$ for all real numbers $ x)$,
the average of all roots is $ a$.[/hide]


Wow that's some high quality BS right there.
And I only read the second paragraph.
[quote]Therefore, almost all Asian children 1) will play an instrument (a dignified classical instrument at that, and certainly not electric guitar)[/quote]
what the... I know plenty of Asians who play electric guitar and is in a band that they made.
[quote]will study math at an accelerated pace[/quote]
I know about 10 Asians who skipped a math class at my school. Certainly that's not a majority of the Asians here.
[quote]will participate in as many extracurricular activities as possible (all academic though, never sports)[/quote]
WHAT THE.....who says Asians never do sports?
[quote]4) take standardized tests early and as many times as necessary to achieve a perfect score.[/quote]
jesus christ I know tons of Asians with Cs and Bs in classes

Half of it is probably false, and the other half is opinion. Who would waste their time to write that... :rotfl:

wtf :censored:....i'm on our school basketball team...how do I not have a social life? i completley agree with bpms.   what kind of ______ would write that? :mad:

Check out this article that was written sometime last year about my school and other schools in our area.

http://wsjclassroom.com/teen/teencenter/05nov_whiteflight.htm

It's true that the majority of the school is Asian, but that's just about the only true thing in that article.  What I think is funny, though, is that the person who wrote the article has an Asian-sounding name.

I saw a book at our local bookstore titled "How to raise your child like an Asian". The same stereotypes apply.

In my case, and in a bunch of my friends' cases, however, the stereotypes are sort of true, and we make no effort to try to correct them (because it gives us ... shall I say ... an elitist feeling or something like that). We also go as far as to say certain academically accomplished non-Asians are "honorary Asians".

Screw PC. Stereotypes are fun.

[quote="Piggypi314"]
http://wsjclassroom.com/teen/teencenter/05nov_whiteflight.htm
[/quote]

Too lazy to take direct quotes from the article but...

First of all, I think the white parents who have a problem with the growing Asian population have several optimal solutions and do not need to complain about it. They feel that the school is centering on academics, and obviously the school is doing so for a good reason. The primary purpose of a school is to provide an education, not to give the students a social life or help them become an NBA All Star. While these are added benefits, the purpose for kids to go to school is to get a good education and get good test scores. In my opinion, if the parents do not care for the increased focus on high test scores, they should unenroll their child, not just from that school, but from all schools in general. Their child would do better to be homeschooled, or not even to be schooled at home, but to just stay home. Perhaps their child, who cares greatly about sports, can join a local travel team or a league or something. That way they can join the Olympics and not have to worry about high test scores, or any test scores for that matter. If the white parents worry so much about paying taxes towards those Asian schools, my advice to them would be to move to Montana; there are a lot of neo-Nazi regions in Montana and they would do better to live there so they wouldn't have to pay for "Asian schools."

As a last comment, I see nothing wrong with a school not having a boy's volleyball team. My school has somewhere around 1800 people, couldn't care less about test scores, has a low percentage of Asian people, and we still don't have a boy's volleyball team.


Hello.

I recently began volunteering for a mentor program where I get to help elementary school kids that don't really have anyone at home who can help with math.  Now, I've tutored my fair share of calculus students and I've been helping people out with problem solving for even longer.  But of all of these, this is definitely the hardest.

What can you say when the kids can't figure out why .6 > .45 or why .25 = .250?  Right now they are learning fractions and decimals.  So I tried to think back to how I learned it and decided to try a number line approach.  We found where 1.5 was, but then 1.7 was placed between 0 and 1.

Now I know that is just a specific example, but I am looking for a more general look at grade school learning.  As much as I would love to deconstruct everything they know and build them back up from the integers, I only meet with these kids twice a week, about one hour each time.

I have learned so much about teaching problem solving, but does anyone know any useful tips for teaching students struggling with the very basics?  Like I know I should try to make it fun and I'm working on some kind of game/reward thing.  But something like an effective (yet easy to obtain) workbook would be great since I don't get much time with them.

I know my thoughts are jumping around a lot, but if you can make sense of what I wrote and offer opinions, I would appreciate it.

Having never taught elementary school math, I cannot offer much. Perhaps I would avoid thinking of "how I learned it", considering that most schoolchildren are not future mloks and their mental processes may be quite different from mine. Of course, volumes have been written on teaching fractions and decimals, and it's not so easy to add something new to them. For example, [url=http://www.fractionbars.com/]Fraction Bars[/url] has a bunch of links along with some content of its own.

Hm, well I hear that AoPS is publishing a new book(s) on teaching elementary-level math soon.  Check the bookstore in a few weeks/months.

[quote="rrusczyk"]We just sent to the printer the first of a series of books we'll be publishing for younger children and their parents.  The first is for parents of very young children (ages 2-8), to give them some direction on teaching them math at home.  We expect to have a second out by the end of the year, and to start publishing a series of workbooks in conjunction with the second book.

The books are written by one of our SDMC instructors, Dr. Chris Wright.  He has a PhD in math from Stanford, yet he still is able to reach younger kids.  The first book should be out in a few weeks.[/quote]

Try to appeal to the kids' interests.  If they are interested in baseball, batting percentages (or whatever it's called, you know what I mean?) are a great way to think about decimals (hitting 0.275 is not as good as/less than hitting 0.388).  I'm not sure how you can relate fractions/decimals to other sports/activities, but I'm sure you can think of something.

One of the best things that a teacher ever told me was that "Those who get it right away are not the ones to be teaching others how to do it."  Not to belittle your efforts - good luck, but try talking to people that had troubles with it (this is probably the wrong place to be then).  Figure out how they mastered decimals, and take ideas from them.  Good luck.  Have fun.  I'll try and think of something else.  As has been mentioned before, there are multitudes of resources on the internet, I'm sure.


I found this link:

http://erdos.fciencias.unam.mx/previousResults.htm

and this one, if you click on the "Warnings and 3-Day Cone",
shows the hurricane headed almost directly at them.

http://www.nhc.noaa.gov

--IMO Dad

[quote="hilbert"]I found this link:

http://erdos.fciencias.unam.mx/previousResults.htm
--IMO Dad[/quote]You should check out the results forum ;) About the hurricane ... everyone here is excited to meet him (her :P).


1)Prove that between any two rational numbers $ {a}$ and $ {b}$  there is at least one other rational number.
2)between two irrational numbers there is at least a rational number
3)Between two rational number is an irrational number.
4)between any two irrational number is an irrational number.

Take the average of the two given rational numbers for 1.

[hide="Solutions?"]1) Consider $ \frac{a\plus{}b}{2}$. It lies between $ a$ and $ b$ and is clearly rational if $ a$ and $ b$ are.

2) Let the two irrational numbers be $ m$ and $ n$, such that $ m>n$. Let $ a$ be the least rational number greater than $ m$, and let $ b$ be the greatest rational number less than $ n$. Clearly, $ \frac{a\plus{}b}{2}\equal{}x$ lies between $ a$ and $ b$, but if it lies outside of $ (m,n)$ then it will contradict the minimality/maximality of $ a$ or $ b$.

3) Consider the decimal representations of the two rational numbers. Clearly, there is some point when they fail to be the same, otherwise, the two will be equal. Then you can construct an irrational number by tacking on a non-terminating, non-repeating sequence of digits to the end of the digit after the last same digits. For instance, given rational numbers $ 3.0$ and $ 3.12$, we would construct $ 3.0\text{non\minus{}repeating sequence}$.

4) Adopting the notation of (2), consider $ \frac{a\plus{}x}{2}$. This number lies between $ a$ and $ x$, and so lies between $ m$ and $ x$ (otherwise contradicting minimality). Then, by (3), we can find an irrational number between this number and $ x$. Then this irrational will lie between $ m$ and $ n$. [/hide]

Construct a non reapeting sequence with two rational number.
E.g 3.0 and 3.12.
@THINK FLOW thanks a lot man.

For instance, $ 3.0141421356\ldots$ would be satisfactory, where $ \sqrt{2}\equal{}1.41421356\ldots$.

[quote="ThinkFlow"]
2) Let the two irrational numbers be $ m$ and $ n$, such that $ m > n$. Let $ a$ be the least rational number greater than $ m$ [/quote]
I don't think you can do this, as the rationals are not well ordered. There is no least rational greater than a given irrational. But you can still use the decimal representation.

Oops, I overlooked that.  :oops:

I guess we can use the decimal representation for all of these problems.

Exactly. The idea behind my solution is that, since $ \sqrt{2}$ (or any irrational number) doesn't have repeating digits, we can concatenate its digits to the end of a rational number in order to make a new irrational number greater than it.

What I did was equivalent to saying that $ 3.0 \plus{} \frac{\sqrt{2}}{100}$ was an irrational number between $ 3.0$ and $ 3.12$.

Between any two real numbers $ a < b$ there exists a rational number $ r$ satisfying $ a < r < b$.

[hide="Proof without using decimal representations"]
WLOG $ a,b$ are both positive. (If they have different signs then 0 is the desired rational number, if they are both negative then replace $ r$ by $ \minus{} r$.) By the Archimedian property of the real numbers there exists a positive integer $ n$ such that $ n > \frac 1{b \minus{} a}$. Hence $ 0 < \frac 1n < b \minus{} a$, and in particular $ b \minus{} \frac 1n > a$. Let $ m$ be the least positive integer such that $ \frac mn > a$. If $ \frac mn\geq b$, then $ \frac {m \minus{} 1}n\geq b \minus{} \frac 1n > a$, which contradicts the minimality of $ m$. Hence $ a < \frac mn < b$, and we are done.[/hide]

[hide="The Archimedian property of the real numbers"]
Let $ x$ be a real number. Then there exists a natural number $ n$ such that $ n > x$.

The proof is straightforward: Suppose $ n\leq x$ for all $ n\in\mathbb N$. Then $ x$ is an upper bound for $ \mathbb N$. By the completeness property of the reals, $ \mathbb N$ also has a [i]least upper bound[/i], which we shall call $ b$. By the definition of $ b$, it follows that $ b \minus{} \frac 12$ is not an upper bound for $ \mathbb N$, in other words, there exists a natural number $ m > b \minus{} \frac 12$, so $ m \plus{} 1 > b \minus{} \frac 12 \plus{} 1 \equal{} b \plus{} \frac 12$, so $ b \plus{} \frac 12$ is not an upper bound for $ \mathbb N$, so neither is $ b$, yielding a contradiction.[/hide]

Faster but more advanced proofs:

[hide="Between every two real numbers there is a rational number"]We'll use the fact that Dedekind cuts are a construction of the reals. Suppose $ a < b$, and suppose for the sake of contradiction that there are no rationals between $ a$ and $ b$ (nonstrictly, i.e., there is no $ q$ such that $ a \leq q \leq b$.) Let $ A \equal{} \{ x \in \mathbb{Q} : x < a \}$ and $ B \equal{} \{x \in \mathbb{Q} : x < b\}$. Since we assumed there are no rationals between $ a$ and $ b$, we get that $ A \equal{} B$. But the Dedekind cut that uniquely corresponds to $ a$ is $ A$, and the Dedekind cut that uniquely corresponds to $ b$ is $ B$, and since $ A \equal{} B$, we have $ a \equal{} b$, a contradiction. [/hide]

[hide="Between every two real numbers there is an irrational number"]Suppose there exists an interval $ (a,b)$ which has no irrational number. It is well known that every interval of reals has the same cardinality as that of the reals, making them uncountable. But by assumption, $ (a,b)$ contains no irrational numbers, so it is a subset of the rationals, which is countable. Thus, $ (a,b)$ is countable and uncountable at the same time, a contradiction.[/hide]

For more information:
[url=http://en.wikipedia.org/wiki/Cardinality]Cardinality[/url]
[url=http://en.wikipedia.org/wiki/Dedekind_cuts]Dedekind cuts, a construction of the reals[/url]


Nitrogen gas reacts with hydrogen gas to form ammonia gas.  You have nitrogen and hydrogen gases in a 15.0-L container fitted with movable piston (the piston allows the container volume to change so as to keep the pressure constant inside the container).  Initially the partial pressure of each reactant gas is 1.00 atm.  Assume the temperature is constant and the reaction goes to completion.

   a)  Calculate the partial pressure of ammonia in the container after the reaction has reached completion.
   b)  Calculate the volume of the container after the reaction has reached completion.

If I can recall correctly, partial pressures are in the same ratio as mole fraction (I think).  Therefore, if you can find the mole ratio at the end of the reaction, that should be equivalent to the partial pressures.

That's what I thought but I am not sure if it is right I did that and I got a partial pressure of 0.67 atm. Is that the right answer???

Problem solved; Thanks!!!

PS: 0.67 atm was not the right answer.

When you say "the reaction goes to completion" does that mean that it goes until the limiting reagent is completely consumed, or that the gas mixture reaches a chemical reaction equilibrium?