suppose a circle.we divide it to n sections. we want to paint each n sections with k colours.calculate the numbers of paintings that we can do so that every two sections does not have the same colour.


limit
logarithms
integration
calculus
ratio
derivative
calculus computations

[b]Find:[/b]

$\lim _{x \rightarrow \infty} \frac{\sum_{k=1}^\infty \frac{(-1)^{k-1}}{k}\frac{x^k}{k!}}{\log{x}+\gamma}$

[b]where gamma is Euler's constant[/b]

We were tying to solve this on another forum. Briefly my thoughts were to show
[hide]
${\lim _ {x \rightarrow \infty} \sum_{k=1}^\infty \frac{(-)^{k-1}}{k}\frac{x^k}{k!}=\gamma +\lim _ {x \rightarrow \infty} \bigg{(} \log x +\int \limits_ {t=1} ^ \infty \frac{e^{-xt}}{t}dt\bigg{)}}$ which was coming from the Puiseux Series for the Exponential Integral, but I have having trobule making the series alternate (breiefly when reapling $x$ by $-x$ in the expanion a logarithm became undefined).

Is this the best way/Is there other ways?[/hide]

[quote="jmerry"]The power series has infinite radius of convergence; that's not a problem. I don't see why you need it to alternate at all.[/quote]

What I meant was the series for Ei(x) has all positive terms, the expression in the numerator of the limit alternates in sign

Obviously, $\gamma$ is irrelevant.

How about L'Hopital:
If the limit of the ratio of derivatives exists,
$\lim_{x\to\infty}\frac{-\sum_{k=1}^\infty\frac{(-x)^k}{k\cdot k!}}{\ln x+\gamma}=\lim_{x\to\infty}\frac{\sum_{k=1}^\infty\frac{(-x)^{k-1}}{k!}}{\frac1x}$
$=\lim_{x\to\infty}\sum_{k=1}^\infty-\frac{(-x)^k}{k!}=\lim_{x\to\infty}1-e^{-x}=1$.

I don't think I've ever done anything quite like that before...


1. In Camberley, England, two out of every seven people have telephone numbers that are not listed in the directory. If there are 14,000 names in the Camberley telephone directory, how many of them have numbers that are unlisted?

2. In Iran, a Westerner cannot take a photograph of a man with a turban. Why not?

3. How many successful parachute jumps does a trainee parachutist in the U.S. Army have to make before he graduates from jump school?

4. What is the invention, first discovered in ancient times, that allows people to see through solid walls?

5. A man carefully pointed his car due east and then drove for two miles. He was then two miles west of where he started from. How come?

6. A mail plane was halfway from Dallas to Miami at a height of 2000 feet on a clear, still day. It dropped a 100 kg sack of airmail letters and a 100 kg steel rod at the same time. Which hit the ground first?

7. If two are company and three are a crowd, what are four and five?

8. With which hand does a nun stir her coffee?

9. Why do some people press elevator buttons with their fingers and others with their thumbs?

10. What do you say to a man who claims not to be superstitious?

11. If a grandfather clock strikes thirteen, what time is it?

12. Two sons, two fathers, and a grandfather sat together. How many men were there?

13. An archaeologist showed his daughter a coin that he had found on a dig. He told her it was dated 200 BC. She told him she thought it was a fake. Who was right and why?

14. What color is a refrigerator?

15. What do cows drink?

Grading Scale
15- Genius
13-14- Smart
11-12- Average
10- Rather Silly
<10- IDIOT

I have the book...hehehe

Can I tell the name of the book?

1. In Camberley, England, two out of every seven people have telephone numbers that are not listed in the directory. If there are 14,000 names in the Camberley telephone directory, how many of them have numbers that are unlisted?

[hide]It is not possible to determine this since many people have multiple phone numbers.
[/hide]
2. In Iran, a Westerner cannot take a photograph of a man with a turban. Why not?

[hide]You take photographs with a camera.
[/hide]
3. How many successful parachute jumps does a trainee parachutist in the U.S. Army have to make before he graduates from jump school?

[hide]The Air Force employs parachutists, not the Army.
[/hide]
4. What is the invention, first discovered in ancient times, that allows people to see through solid walls?

[hide]The window
[/hide]
5. A man carefully pointed his car due east and then drove for two miles. He was then two miles west of where he started from. How come?

[hide]He drove backwards.
[/hide]
6. A mail plane was halfway from Dallas to Miami at a height of 2000 feet on a clear, still day. It dropped a 100 kg sack of airmail letters and a 100 kg steel rod at the same time. Which hit the ground first?

[hide]Neither; both fell into the sea.
[/hide]
7. If two are company and three are a crowd, what are four and five?

[hide]Numbers? Words?
[/hide]
8. With which hand does a nun stir her coffee?

[hide]You don't stir coffee.
[/hide]
9. Why do some people press elevator buttons with their fingers and others with their thumbs?

[hide]It's easier to press the elevator buttons with your fingers, while other buttons may be more easily pressed with your thumb.
[/hide]
10. What do you say to a man who claims not to be superstitious?

[hide]I would probably say "OK."
[/hide]
11. If a grandfather clock strikes thirteen, what time is it?

[hide]Time to get a new clock?
[/hide]
12. Two sons, two fathers, and a grandfather sat together. How many men were there?

[hide]Two.
[/hide]
13. An archaeologist showed his daughter a coin that he had found on a dig. He told her it was dated 200 BC. She told him she thought it was a fake. Who was right and why?

[hide]The daughter was. People in 200 BC did not base their calendar on Christrianity.
[/hide]
14. What color is a refrigerator?

[hide]Refrigerators don't have to be any particular color. I don't know of any color that refrigerates.
[/hide]
15. What do cows drink?

[hide]Water
[/hide]

I think your answers for 1 and 12 are wrong.

#1 should be 0.
If they're in the telephone book, their number is listed.
#9 thumb is a finger?
#12 can be 3, but is a boy a man?

The only one I got correct was number 1. It has to be zero, because don't they only list the people that have telephone numbers?

All men are sons, therefore there were no more than two men.

[hide="My #12 explanation"]All men are sons, but not all men are grandfathers.  Because the question mentions a grandfather, there must be at least two other generations of men.  There are 3 men (who I'll call A, B, and C).  A is father to B and grandfather to C, B is a son of A and father of C, and C is the son of B.  Thus, there are two sons (B and C), two fathers (A and B) and one grandfather (A).[/hide]

That's correct cf249.
It's 3. Nice explanation, although this is a idiot test.

*peered at the top of the thread and examined the grading rubric*

Apparently I'm an idiot. :fool:

We knew [hide]Joke![/hide].
I'm smart with 14 answers :D

1. In Camberley, England, two out of every seven people have telephone numbers that are not listed in the directory. If there are 14,000 names in the Camberley telephone directory, how many of them have numbers that are unlisted?
[hide="answer"]0. If they are in the direstory, all of them are listed.[/hide]

2. In Iran, a Westerner cannot take a photograph of a man with a turban. Why not?
[hide="answer"]You take photographs with a camera, not a turban.[/hide]

3. How many successful parachute jumps does a trainee parachutist in the U.S. Army have to make before he graduates from jump school?
[hide="answer"]None. The U.S. Air Force hires parachutists.[/hide]

4. What is the invention, first discovered in ancient times, that allows people to see through solid walls?
[hide="answer"]The Window.[/hide]

5. A man carefully pointed his car due east and then drove for two miles. He was then two miles west of where he started from. How come?
[hide="answer"]He drove in reverse.[/hide]

6. A mail plane was halfway from Dallas to Miami at a height of 2000 feet on a clear, still day. It dropped a 100 kg sack of airmail letters and a 100 kg steel rod at the same time. Which hit the ground first?
[hide="answer"]Neither, both hit the water.[/hide]

7. If two are company and three are a crowd, what are four and five?
[hide="answer"]9.[/hide]

8. With which hand does a nun stir her coffee?
[hide="answer"]You don't stir coffee.[/hide]

9. Why do some people press elevator buttons with their fingers and others with their thumbs?
[hide="answer"]To get to their floor.[/hide]

10. What do you say to a man who claims not to be superstitious?
[hide="answer"]That's bad luck? I don't know.[/hide]

11. If a grandfather clock strikes thirteen, what time is it?
[hide="answer"]A grandfather clock can only strike up to 12.[/hide]

12. Two sons, two fathers, and a grandfather sat together. How many men were there?
[hide="answer"]3.[/hide]

13. An archaeologist showed his daughter a coin that he had found on a dig. He told her it was dated 200 BC. She told him she thought it was a fake. Who was right and why?
[hide="answer"]She was, because people living in such-and-such B.C. wouldn't date themselves B.C. because Christ hadn't come yet.[/hide]

14. What color is a refrigerator?
[hide="answer"]Isn't it "refrid[b]g[/b]erator"?[/hide]

15. What do cows drink?
[hide="answer"]Water. What else?[/hide][/hide]

Actually, erabel, "refrigerator" is spelled correctly.

For #15, some people's first answer would be:
[hide]milk.   :rotfl: [/hide]

@erabel - I do not think that 10 and 11 are correct. I think fourteen was just a typo.

@mathlete Is the scientific answer to #5 [hide="this?"]The two objects have the same mass because they were weighed under the same gravity, and there is little air resistance on a clear, brisk day. Thus, the objects will hit the sea at the same time.[/hide]

[quote="Wickedestjr"]The third idiot test makes no sense. It says click on the largest rectangle. Can somebody help me on this one?[/quote]
Click on the border of the test window.

When you pass the Idiot Test 5, it calls you the Ultimate Idiot. ???
BTW, I passed Idiot Test 3

Yay!!! I finally beat the third idiot test. When it asked a question similar to that one, I clicked on the border, but it didn't do anything, so I continued to believe it would never work.

couldn't the answer to #11 also be 'time to get a new clock'?

on another topic, who here has beaten/take the impossible quiz1(2). i've gotten to q 100 on 1 and then lost, and i don'ot remember record for 2

2 is alot harder 'cause you can't cheat as easily.

alternate solution no.6

[hide]
he's so close to the south pole, 2 miles east is half way around the world!

[/hide]

I think the reasoning behind number 10 is this:

The man who is being asked to lend out 13 dollars will refuse, because the two men are strangers. Thus, the questioner retorts by saying he is superstitious for not lending out 13 dollars.

For number four,  I don't see why it can't be a hole... or maybe something like a piece of glass stuck in the wall...

What's a hole in the wall with a piece of glass in it?  :D

Well then why not a hole? Why so specific and be a window? I mean it's just a hole that's not covered up in the middle of the wall...

For the window problem: Well, If it was a hole or window in the wall, then you would be seeing through the window (or the hole) instead of the wall.


$(1-x^{2})y''-xy'+y=0$ is the Cebisev differential equation with n=1, so y=x in virtute that the $(1-x^{2})y''-xy'+n^{2}y=0$ has its solutions $T_{n}(x)=\frac{1}{2^{n-1}}cos(n arccos x)$

Note that $y=x$ is a particular solution. Now look for solutions of de form $y=xv$


Let $ f(x)\equal{}1\plus{}2\cos x\plus{}3\sin x$. Find the values of $ a,\ b,\ c$ such that $ af(x)\plus{}bf(x\minus{}c)\equal{}1$ for any real numbers $ x$.

hello, it is $ a\equal{}\frac{1}{2},b\equal{}\frac{1}{2},c\equal{}0$.
Sonnhard.

[hide="Hint"]For $ a\equal{}b\equal{}{1\over 2},c\equal{}\pi$ the equality is satisfied, but is that the only possible value of $ c$?[/hide]

The question does not specify to find all possible values for $ a, b, c.$  Right now is a matter of proving $ a \equal{} b \equal{} 1/2$ and $ c \equal{} (2n\plus{}1)\pi, n \in \mathbb{Z}$ are the ONLY possible solutions, if it is indeed the case.

[hide="Second hint"] Let $ \alpha \equal{} \arctan \frac{3}{2}$ and note that $ f(x) \equal{} 1 \plus{} \sqrt{13} \cos (x \minus{} \alpha)$.  What restrictions does this place on $ a, b$? [/hide]


trigonometry
inequalities
inequalities proposed

Thank you for your nice solution. if we make use of $xy+yz+zx=1$ where $x=\tan (A/2), y=\tan (B/2)$ and $z=\tan (C/2)$, $A, B, C$ are angles of a triangle. We can also solve the problem. Specifically, the desired inequality is equivalently written as
\[\cos A+\cos B+\cos C-4\cos A\cos B\cos C\geq 1\]
The rest is for you.

[quote="yptsoi"]:)  I just write blahblahblah's method in more details:

Let $a,b,c$ be $\cot\frac{A}{2},\cot\frac{B}{2},\cot\frac{C}{2}$ respectively where $A+B+C=180^{\cdot}$, then the given condition has been satisfied.
Using the two identities
\[1+cot^2x=csc^2x,\]
\[\sin\frac{A}{2}+\sin\frac{B}{2}+\sin\frac{C}{2} = 4\sin\frac{A+B}{4}\sin\frac{B+C}{4}\sin\frac{C+A}{4}+1\]
, the inequality can be simplified to the following equivalent form:
\[\csc\frac{A}{2}\csc\frac{B}{2}+\csc\frac{B}{2}\csc\frac{C}{2}+\csc\frac{C}{2}\csc\frac{A}{2} \ge 4+ \csc\frac{A}{2}\csc\frac{B}{2}\csc\frac{C}{2} \Leftrightarrow\]
\[\sin\frac{A}{2}+\sin\frac{B}{2}+\sin\frac{C}{2} \ge 4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}+1 \Leftrightarrow\]
\[\sin\frac{A+B}{4}\sin\frac{B+C}{4}\sin\frac{C+A}{4} \ge \sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}.\]
By AM-GM inequality,
\[\sin\frac{A+B}{4} = \sin\frac{A}{4}\cos\frac{B}{4}+\cos\frac{A}{4}\sin\frac{B}{4} \ge \sqrt{4\sin\frac{A}{4}\cos\frac{A}{4}\sin\frac{B}{4}\cos\frac{B}{4}}=\sqrt{\sin\frac{A}{2}\sin\frac{B}{2}},\]
and similarly we have $\sin\frac{B+C}{4} \ge \sqrt{\sin\frac{B}{2}\sin\frac{C}{2}}$ and $\sin\frac{C+A}{4} \ge \sqrt{\sin\frac{C}{2}\sin\frac{A}{2}}$.
Then multipling all the terms on R.H.S and L.H.S respectively gives the inequality.[/quote]

My solution is similary  :lol:

[quote]
Using the identity,
$1+x^{2}=1+x^{2}\frac{x+y+z}{xyz}=\frac{(x+y)(x+z)}{yz}$,
the desired inequality is equivalent to
$\sum\sqrt{(1+y^{2})(1+z^{2})}-4-\sqrt{\prod(1+x^{2})}$
$=\sum\frac{y+z}{x}\sqrt{\frac{(x+y)(x+z)}{yz}}-4-\prod\frac{y+z}{x}$
$=\sum{\frac{(y+z)\sqrt{yz(x+y)(x+z)}-yz(2x+y+z)}{xyz}}$
$=\sum\frac{(x+y+z)(y-z)^{2}}{(y+z)\sqrt{yz(x+y)(x+z)}+yz(2x+y+z)}\geq0$.
[/quote]

I'm very suprised about S.O.S method can solve this problem

[quote="Ji Chen"]
Using the identity,
$1+x^{2}=1+x^{2}\frac{x+y+z}{xyz}=\frac{(x+y)(x+z)}{yz}$,
the desired inequality is equivalent to
$\sum\sqrt{(1+y^{2})(1+z^{2})}-4-\sqrt{\prod(1+x^{2})}$
$=\sum\frac{y+z}{x}\sqrt{\frac{(x+y)(x+z)}{yz}}-4-\prod\frac{y+z}{x}$
$=\sum{\frac{(y+z)\sqrt{yz(x+y)(x+z)}-yz(2x+y+z)}{xyz}}$
$=\sum\frac{(x+y+z)(y-z)^{2}}{(y+z)\sqrt{yz(x+y)(x+z)}+yz(2x+y+z)}\geq0$.

[hide="New problem."]
If $x,y,z$ are positive real numbers such that $x+y+z=xyz$, then
$34+\sum\sqrt{(1+34y^{2})(1+34z^{2})}\geq \sqrt{\prod(1+16x^{2})}\geq 205+\sum\sqrt{(1+15y^{2})(1+15z^{2})}$.
[/hide][/quote]
Very nice solution, Ji Chen.

Thank hien and pvthuan. Here is another new problem:
If $x,y,z$ are positive real numbers such that $x+y+z=xyz$, then
$\sum\sqrt{(11+y^{2})(11+z^{2})}\leq 15+\sqrt{\prod(6+x^{2})}$.


By Holder's inequality

$\prod (1+a_i)\geq \left(1+\sqrt[n]{\prod a_i}\right)^n \geq 1+\sqrt[n]{\prod a_i}$


trigonometry
trig identities
tangent half angle

i cannot prove this no matter what i try to do! I need help to be able to pass my test tommorrow.

Prove sin(t)/(1-cos(t)) = (1+cos(t)) / sin(t)

I worked with the left terms:

sin(t) / (1-cos(t)) | Problem

sin(t) / (sin^2(t) + cos^2(t)-cos(t)) | Pythagorean ID:sin^2(t) + cos^2(t) = 1

what to do next? i already tried seperating all the terms and that doesn't work out. I have a test tomorrow on this stuff and i really need some help.

none of the moderators are on.... i just checked. *Boo Hoo* :(

I would start with:
$sin^2(t)+cos^2(t)=1$
subtracting $cos^2(t)$ from bothe sides we obtain $sin^2(t)=1-cos^2(t)$
Now factor both sides as follows $(sin(t))(sin(t))=(1-cos(t))(1+cos(t))$
Next divide both sides by $(sin(t))(1-cos(t))$
this produces $(sin(t))/(1-cos(t))=(1+cos(t))/(sin(t))$
Now we have proved that the desired expression holds true as long as sin(t) doesn't equal 0.

[hide]
Just multiply the LHS by $\frac{1+\cos{t}}{1+\cos{t}}$ and see what happens ;) [/hide]

I'm a moderator!  :)

If you remember, those are both the half-angle formulas for a tangent. Just reverse the way you derive those tangent half-angle formulas. In that case, that step would be...

[hide="Think about it first"]$\frac{1+ \cos t}{1+ \cos t}$ I think that this should'be veen in Intermediate cause I can solve it.  ;) [/hide]

I'm a moderator.  But I probably wouldn't have solved it anyways (sorry, but it's the truth...I'm busy!).  Btw, I moved this to intermediate.


What is your favorite kind of candy? Mine is Three Musketeers!

I like skittles, m&ms, kit-kats, herscheys's, and reese's. I like a lot more too. Just too long of a list. :D

The Instant Noodles, Pretzels, and Pop Secret Popcorn I got were really good. Don't really like sweet things.

Mike and IKes all the way.

Skittles and 3 muskers also rock.

I like almost everything chocolate. :blush:

[quote="mustafa"]What is werthers? I've never heard of it before.[/quote]

wait. you never tried [url=http://www.storck.com/de/brand/werthers/]werthers[/url]
you must try it. it will change you.

[quote="Nienna"]I like almost everything chocolate. Blush[/quote]

me too. the higher the percentage of chocolate the sweeter (well, not always the better).
and how much i eat chocolate before, during and after the math and physics competitions.  :oops:
probably, the world would be very different without the chocolate and candies
(but [b]the chocolate[/b] is stressed).  :oops:  :P  :oops:

Gotta be junior mints all the way.  I'm addicted.

ANYTHING WITH SUGAR THAT IS SWEET! AND ANYTHING SOUR THAT IS ASSOCIATED WITH CANDY!

  :rotfl:

Everything with chocolate is good, but my favorite...CRUNCH BARS!

CHOCOLATE!!!! :first:

and sour candy is pretty good

I just tried chocolate with amaretto filling. It's beautiful and this will be nice winter. :yup:

[quote="mustafa"]What is werthers? I've never heard of it before.[/quote]

suckable caramel.

[quote="now a ranger"]caramel is awesome

and i cant eat it anymore ever since I got my braces on two weeks ago  :maybe:[/quote]

Lol you actually listen to your orthodontist?

twix, kit-kat, crunch bars, three musketeers, skittles, gummy worms (non-sour)


trigonometry
complex numbers
algebra theorems
algebra

You can take it as definition, but more common (and better for analysis) is to look at the Taylor series of the desired funvtion:
$e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$
$\sin(x) = \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!}$
$\cos(x) = \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n)!}$

Thank you,ZetaX.
And,what is the use of it ?

There should be a lot of examples, like:
At first, the identity $(e^{ix})^n = e^{ixn}$ looks trivial (it isn't totally trivial since some rules of exponentiation only hold for (positive) real numbers).
But when you write this identity as $(\cos(x)+i \cdot \sin(x))^n = \cos(nx) + i \cdot \sin (nx)$ and consider that two complex numbers are equal iff the real and the imaginery part are so, then you get a formula for $\sin(nx) , \cos(nx)$ only using $\sin(x),\cos(x)$.


geometry
projective geometry
power of a point
radical axis
geometry proposed

$ D.E.F$ are the midpoints of $ BC,CA,AB$
$ DL,EN,FM$ are tangent to the inscribecircle of $ ABC$
$ L\in EF$
$ M\in DE$
$ N\in DF$
Prove that $ L,M,N$ are collinear

[b]Lemma:[/b] Given triangle $ ABC$ and circle $ (O). (O)\cap BC, CA, AB \equal{} \{A_1,A_2,B_1,B_2,C_1,C_2\}$. Then $ AA_1,BB_1, CC_1$ are concurrent iff $ AA_2,BB_2,CC_2$ are concurrent.
[b]Proof:[/b] We can prove this lemma easily by using power of a point and Ceva's theorem.
[b]Return to our problem:[/b]
See the names of points on the figure.
A theorem about circumscribed quadrilateral tells us $ ZB_1, YC_1, EF$ concur at $ R$.
Denote $ A' \equal{} C_1Z\cap B_1Y. (EFR)$ is the polar of $ A'$ wrt $ (I)$ then $ A'I\perp EF$ or $ A',I,A_1$ are collinear.
Similarly we get $ A'A_1, B'B_1, C'C_1$ are concurrent.
Applying the lemma above we get $ A'X, B'Y, C'Z$ concur at $ K$.
$ L$ lies on the polar of $ A'$ then $ A'$ lies on the polar of $ L$. Therefore $ A'X$ is the polar of $ L$.
$ \Rightarrow L$ lies on the polar of $ K$.
Similarly we obtain $ L,M,N$ lie on the polar of $ K$. Our proof is completed then.
[b]Remark:[/b] $ AX, BY, CZ$ concur at [b]Nagel [/b]point of triangle $ ABC$.

Let $ X,Y,Z$ be the tangency points of $ (I)$ with $ BC,CA,AB$ and $ D',E',F'$ the contact points of the tangents from $ D,E,F$ to $ (I),$ different from $ X,Y,Z.$ The poles $ A',B',C'$ of $ EF,FD,DE$ WRT $ (I)$ lie on $ IX,IY,IZ,$ respectively due to $ EF \parallel BC,$ $FD \parallel AC$ and $DE \parallel AB.$ Thus, polars of $ D,E,F$ WRT $ (I),$ which are the lines $ B'C',C'A',A'B',$ are identical with $ XD',$ $YE',$ $ZF'$ $ \Longrightarrow$ $ L,N,M$ are the poles of $ A'D',B'E',C'F'.$ Since $ A'X,B'Y$ and $C'Z$ concur at $ I,$ then $ A'D',$ $B'E'$ and $C'F'$ concur at the cyclocevian $ I'$ of $ I$ in $ \triangle A'B'C' \cup (I).$ Therefore, $ L,N,M$ are collinear on the polar of $ I'$ WRT $ (I).$

$ \bullet$ Another result coming from this configuration:

$ \triangle ABC$ is scalene and let $ X,Y,Z$ be the tangency points of the excircles $ (I_a),(I_b),(I_c)$ with $ BC,CA,AB.$ Perpendicular bisectors of $ AX,BY,CZ$ meet $ BC,CA,AB$ at $ A',B',C',$ respectively. Then $ A',B',C'$ are collinear.

This nice problem, has already been posted before, as [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=140191]Interesting and hard[/url], by [b]cvix[/b].

[quote="livetolove212"][b]Lemma:[/b] Given triangle $ ABC$ and circle $ (O). (O)\cap BC, CA, AB \equal{} \{A_1,A_2,B_1,B_2,C_1,C_2\}$. Then $ AA_1,BB_1, CC_1$ are concurrent iff $ AA_2,BB_2,CC_2$ are concurrent.[/quote]
There are some ( not well known I think ) interesting concurrencies and one collinearity, in the configuration of an arbitrary circle $ (K)$, cutting the side-segments $ BC,\ AC,\ AB$ of a given triangle $ \bigtriangleup ABC,$ at pairs of points $ D,\ D'$ and $ E,\ E'$ and $ F,\ F',$ respectively ( the points $ D,\ E,\ F$ are adjoining to the vertices $ B,\ C,\ A,$ ).

[b][size=100][color=DarkBlue]$ \bullet$ The line segments $ AA',\ BB',\ CC',$ are concurrent at one point so be it $ P,$ where $ A'\equiv BE\cap CF'$ and $ B'\equiv AD'\cap CF$ and $ C'\equiv AD\cap BE'$.

$ \bullet$ The line segments $ AA'',\ BB'',\ CC'',$ are concurrent at one point so be it $ Q,$ where $ A''\equiv BE'\cap CF$ and $ B''\equiv AD\cap CF'$ and $ C''\equiv AD'\cap BE$.

$ \bullet$ The line segments $ A'A'',\ B'B'',\ C'C'',$ are concurrent at one point so be it $ R,$ lying on the line segment $ PQ.$[/color][/size][/b]

Best regards, Kostas Vittas.

Thank you for precious comment :lol:
I paste four different solutions below(very sorry if repeated)
The first is purely algebraic(I like calculating all the time :D )
The third used [b]Brianchon theorem[/b]
The second and the last based on the theorem of radical axis
Wish you like them!

Here is my solution.

Let $ A_0$ be the point of tangency of $ (I)$ and $ BC$, $ A_0A_3$ is the diameter of $ (I)$, $ A_0' \equiv AA_3 \cap BC$, $ A_5 \equiv AA_3 \cap B_1C_1$ and $ A_4$ is the point of tangency of $ (I)$ and $ DL$. It is known that $ A,A_3,A_5,A'_0$ are collinear. We have $ DA_0 \equal{} DA_4 \equal{} DA_0'$, which yields $ A_0A_4 \bot A_4A'_0$. On the other hand, $ DI \bot A_0A_4$ and $ D,I$ are the midpoints of $ A_0A'_0$ and $ A_0A_3$ so $ DI \parallel A_3A'_0$. Hence $ A_3A'_0 \bot A_0A_4$. Therefore $ A_4 \in \overline{AA_3A_5A'_0}$. Moreover, $ \angle LA_5A_4 \equal{} \angle DA'_0A_4 \equal{} \angle DA_4A'_0 \equal{} \angle LA_4A_4$ so $ LA_4^2 \equal{} LA_5^2$, meanwhile $ L$ lies on the radical axis of $ (I)$ and $ (I')$. Use the same argument of $ M,N$, we have $ L,M,N$ are collinear since they are all belong to the radical axis of $ (I)$ and $ (I')$, QED.

[b]Generalized problem[/b] (Stated by PDatK40SP)
Let $ ABC$ be a triangle with incircle $ (I)$ and $ A_1,B_1,C_1$ are three arbitrary points on $ BC,CA,AB$. Denoted by $ A_2$ the intersection of $ B_1C_1$ with the tangent of $ (I)$ from $ A_1$ but different of $ BC$. Prove that the $ A_2,B_2,C_2$ are collinear iff $ AA_1,BB_1,CC_1$ concur.

[quote="vittasko"]
There are some ( not well known I think ) interesting concurrencies and one collinearity, in the configuration of an arbitrary circle $ (K)$, cutting the side-segments $ BC,\ AC,\ AB$ of a given triangle $ \bigtriangleup ABC,$ at pairs of points $ D,\ D'$ and $ E,\ E'$ and $ F,\ F',$ respectively ( the points $ D,\ E,\ F$ are adjoining to the vertices $ B,\ C,\ A,$ ).

[b][size=100][color=DarkBlue]$ \bullet$ The line segments $ AA',\ BB',\ CC',$ are concurrent at one point so be it $ P,$ where $ A'\equiv BE\cap CF'$ and $ B'\equiv AD'\cap CF$ and $ C'\equiv AD\cap BE'$.

$ \bullet$ The line segments $ AA'',\ BB'',\ CC'',$ are concurrent at one point so be it $ Q,$ where $ A''\equiv BE'\cap CF$ and $ B''\equiv AD\cap CF'$ and $ C''\equiv AD'\cap BE$.

$ \bullet$ The line segments $ A'A'',\ B'B'',\ C'C'',$ are concurrent at one point so be it $ R,$ lying on the line segment $ PQ.$[/color][/size][/b]

Best regards, Kostas Vittas.[/quote]

Dear [b]Kostas[/b],
I have other similar concurrencies.
Given triangle $ ABC.(O) \cap BC, CA, AB\equal{}\{A_1,A_2,B_1,B_2,C_1,C_2\}. X_1\equal{}A_1B_1\cap A_2C_2$, similar for $ Y_1, Z_1. X_2\equal{}C_2B_2\cap C_1B_1$, similar for $ Y_2, Z_2. A_3\equal{}A_1C_1\cap A_2B_2$, similar for $ B_3, C_3$. Prove that:
a, $ X_1X_2, Y_1Y_2, Z_1Z_2$ concur at $ D$.
b, $ A_3X_1, B_3Y_1, C_3Z_1$ concur at $ E$.
c, $ A_3X_2, B_3Y_2, C_3Z_2$ concur at $ F$.
d, $ D,E,F$ are collinear.

[quote="livetolove212"]Given triangle $ ABC.(O) \cap BC, CA, AB \equal{} \{A_1,A_2,B_1,B_2,C_1,C_2\}. X_1 \equal{} A_1B_1\cap A_2C_2$, similar for $ Y_1, Z_1. X_2 \equal{} C_2B_2\cap C_1B_1$, similar for $ Y_2, Z_2. A_3 \equal{} A_1C_1\cap A_2B_2$, similar for $ B_3, C_3$. Prove that:
a, $ X_1X_2, Y_1Y_2, Z_1Z_2$ concur at $ D$.
b, $ A_3X_1, B_3Y_1, C_3Z_1$ concur at $ E$.
c, $ A_3X_2, B_3Y_2, C_3Z_2$ concur at $ F$.
d, $ D,E,F$ are collinear.[/quote]
Thank you dear [b]Linh[/b] for another group of familiar concurrencies and collinearity, new to me.

I think that the key of the solution is the collinearity of $ A_{3},\ A,\ X_{1}$ ( similarly of the triads $ B_{3},\ B,\ Y_{1}$ and $ C_{3},\ C,\ Z_{1}$ ) and the concurrency of $ F\equiv A_{3}X_{2}\cap B_{3}Y_{2}\cap C_{3}Z_{2}$.

We can after to complete the proof, applying the [b]Desmic theorem[/b].

I will check my notes and if I am not mistaken, I will post here later more deatails ( it is a crazy dancing of the [b]Pascal[/b] and [b]Desarques[/b], theorems   :)  ).

Best regards, Kostas Vittas.

Dear [b]Kostas[/b],
I found an interesting concurrency:
Denote $ A_4\equal{}BB_2\cap CC_1, A_5\equal{}BB_1\cap CC_2$, similar for $ B_4,B_5,C_4,C_5$. Prove that 6 lines $ AX_1, BY_1, CZ_1, A_4A_5, B_4B_5, C_4C_5$ concur at $ E$.

[quote="livetolove212"] Given triangle $ ABC.(O) \cap BC, CA, AB \equal{} \{A_1,A_2,B_1,B_2,C_1,C_2\}. X_1 \equal{} A_1B_1\cap A_2C_2$, similar for $ Y_1, Z_1. X_2 \equal{} C_2B_2\cap C_1B_1$, similar for $ Y_2, Z_2. A_3 \equal{} A_1C_1\cap A_2B_2$, similar for $ B_3, C_3$. Prove that:
a, $ X_1X_2, Y_1Y_2, Z_1Z_2$ concur at $ D$.
b, $ A_3X_1, B_3Y_1, C_3Z_1$ concur at $ E$.
c, $ A_3X_2, B_3Y_2, C_3Z_2$ concur at $ F$.
d, $ D,E,F$ are collinear.[/quote]
$ \bullet$ Applying the [b]Pascal theorem[/b] in cyclic hexagon $ A_{1}A_{2}B_{1}B_{2}C_{1}C_{2},$ we have that the points $ P\equiv A_{1}A_{2}\cap B_{2}C_{1}$ and $ Q\equiv B_{1}B_{2}\cap A_{1}C_{2}$ and $ R\equiv A_{2}B_{1}\cap C_{1}C_{2},$ are collinear.

Applying again the [b]Pascal theorem[/b] in cyclic non-convex hexagon $ A_{1}C_{1}C_{2}B_{2}A_{2}B_{1},$ we have that the points $ B_{3}\equiv C_{2}B_{2}\cap B_{1}A_{1}$ and $ A_{3}\equiv A_{1}C_{1}\cap B_{2}A_{2}$ and $ R\equiv C_{1}C_{2}\cap A_{2}B_{1},$ are collinear and similarly, we have the collinearities of the points $ C_{3},\ B_{3},\ P$ and $ C_{3},\ A_{3},\ Q.$

$ \bullet$ Let be the points $ A'\equiv A_{2}B_{1}\cap A_{1}C_{2},\ B'\equiv A_{2}B_{1}\cap B_{2}C_{1},\ C'\equiv A_{1}C_{2}\cap B_{2}C_{1}$ and because of the collinearity of the points $ P,\ Q,\ R,$ based on the [b]Desarques theorem[/b], we conclude that the triangles $ \bigtriangleup A_{3}B_{3}C_{3},\ \bigtriangleup A'B'C'$ are perspective and then, we have that $ F\equiv A_{3}A'\cap B_{3}B'\cap C_{3}C'$ $ ,(1)$

We consider now, the cyclic non-convex hexagon $ A_{1}C_{1}B_{1}A_{2}B_{2}C_{2},$ and based on the [b]Pascal theorem[/b], we have that the collinearity of the points $ A_{3}\equiv A_{1}C_{1}\cap A_{2}B_{2}$ and $ X_{2}\equiv C_{1}B_{1}\cap B_{2}C_{2}$ and $ A'\equiv B_{1}A_{2}\cap C_{2}A_{1},$ are collinear.

That is, the point $ X_{2},$ lies on the line segment $ A_{3}A'$ and similarly, the points $ Y_{2},\ Z_{2},$ lie on the line segments $ B_{3}B',\ C_{3}C',$ respectively.

$ \bullet$ So, we have now the configuration of the concurrent segments $ A_{3}X_{2},\ B_{3}Y_{2},\ C_{3}Z_{2}$ and based on the [b]Desmic theorem[/b] ( it is what I have called as [b]Second theorem of concurrent segments[/b] ), we conclude the concurrencies of  $ E\equiv A_{3}X_{1}\cap B_{3}Y_{1}\cap C_{3}Z_{1}$ and $ D\equiv X_{1}X_{2}\cap Y_{1}Y_{2}\cap Z_{1}Z_{2}$ and the collinearity of the points $ F,\ D,\ E$.

Applying last time [b]Pascal theorem[/b] in the cyclic non-convex hexagon $ A_{1}C_{1}C_{2}A_{2}B_{2}B_{1},$ we have the collinearity of the points $ A_{3},\ A,\ X_{1}.$

That is, the point $ A$ lies on the line segment $ A_{3}X_{1}$ and similarly, the points $ B,\ C,$ lie on the line segments $ B_{3}Y_{1},\ C_{3}Z_{1}$ and the proof is completed.

Kostas Vittas.

[quote="livetolove212"]Denote $ A_4 \equal{} BB_2\cap CC_1, A_5 \equal{} BB_1\cap CC_2$, similar for $ B_4,B_5,C_4,C_5$. Prove that 6 lines $ AX_1, BY_1, CZ_1, A_4A_5, B_4B_5, C_4C_5$ concur at $ E$.[/quote]
It has been posted before in Hyacinthos Forum, by [b]Neagoe Petrisor[/b].

Please see at:
http://tech.groups.yahoo.com/group/Hyacinthos/message/14694?threaded=1&l=1

http://f1.grp.yahoofs.com/v1/8I7qSiVX4nPCMuMD9Ia0jLzuvgh_5iLIK3e6TvCc2JkLHDX2p4jDWN4tTtKOM7qanEJLECNHGDt-B8zKNmV33rXnP_GV/concurent%20line.pdf

Best regards, Kostas Vittas.

[quote="Mashimaru"][b]Generalized problem[/b] (Stated by PDatK40SP) Let $ ABC$ be a triangle with incircle $ (I)$ and $ A_1,B_1,C_1$ are three arbitrary points on $ BC,CA,AB.$ Denoted by $ A_2$ the intersection of $ B_1C_1$ with the tangent of $ (I)$ from $ A_1$ but different of $ BC.$ Prove that the $ A_2,B_2,C_2$ are collinear iff $ AA_1,BB_1,CC_1$ concur.[/quote]
Let $ D,E,F$ be the tangency points of $ (I)$ with $ BC,CA,AB$ and $ D',E',F'$ the contact points of the tangents from $ A_1,B_1,C_1$ to $ (I),$ different from $ D,E,F.$ Poles $ A_0,B_0,C_0$ of the lines $ B_1C_1,C_1A_1,A_1B_1$ WRT $ (I)$ form a triangle circumscribed in $ \triangle DEF$ and $ \triangle D'E'F',$ since $ DD',EE',FF'$ are identical to the polars of $ A_1,B_1,C_1$ WRT $ (I)$ $ \Longrightarrow$ $ A_0D',B_0E'$ and $ C_0F'$ are the polars of $ A_2,B_2,C_2.$

On the other hand, $ DA_0,EB_0,FC_0$ are polars of $ A' \equiv BC \cap B_1C_1,$  $ \ B' \equiv CA \cap C_1A_1$ and $ C' \equiv AB \cap A_1B_1.$ By Desargues theorem, the points $ A',B',C'$ are collinear if and only if $ AA_1,BB_1,CC_1$ concur at a point $ P$ $ \Longrightarrow$ $ A_0D,B_0E$ and $C_0F$  concur at the pole $Q$ of the trilinear polar  of $ P$ WRT $(I).$  Thus, $ A_0D',B_0E'$ and $ C_0F'$ concur at the cyclocevian $Q'$ of $ Q$ WRT $ \triangle A_0B_0C_0$ $ \Longrightarrow$  $ A_2,B_2,C_2$ lie on the polar of $ Q'$ WRT $ (I).$


[hide="Solution"]We have
3x^2 + 4x - 44 = 20
3x^2 + 4x - 64 = 0 We factor this to get
(3x + 16)(x -4) = 0 Using the Zero Product Property, we get
3x + 16 = 0 or x - 4 = 0

[b]x = -16/3, 4[/b][/hide]


Two circles with radii of $ 2$ and $ 4$, and centers $ A$ and $ B$, respectively, are externally tangent. A third circle is externally tangent to circles $ A$ and $ B$ has a center $ C$. Let $ M$ be the midpoint of $ AB$. If $ \angle AMC\equal{}\theta$ is at a maximum, find $ \tan \theta$.

[hide="what i got so far"]$ CM\equal{}x$. r is the radius of circle C.
$ (4\plus{}r)^2\equal{}3^2\plus{}x^2\plus{}6x\cos\theta$
$ (2\plus{}r)^2\equal{}3^2\plus{}x^2\minus{}6x\cos\theta$

$ \implies x^2\equal{}\frac{(4\plus{}r)^2\plus{}(2\plus{}r)^2\minus{}2\cdot3^2}2$

$ \implies (4\plus{}r)^2\equal{}(2\plus{}r)^2\plus{}12x\cos\theta$

$ \implies \cos\theta\equal{}\frac{(4\plus{}r)^2\minus{}(2\plus{}r)^2}{12x}$

Now it seems really messy to find the extrema (x in terms of r).[/hide]

Write cos in terms of tan and then simplify it out.

Does anyone have any idea on finding the extrema in a less computational way?

[hide="I think"]
Is the answer $ \tan \theta \equal{} \frac {8\sqrt {5}}{5}$?
If so...
Call $ X$ the point of tangency of the two circles with radii $ 2$ and $ 4$. Then as the radius of the circle with center $ C$ gets infinitely large, $ \theta \equal{} \angle AMC$ becomes $ \angle AXC$, which is its maximum (I cannot prove this but it seems intuitive). Then you find $ \angle AXC \equal{} \theta$ using stewarts theorem and then calculate $ \tan\theta$.
[/hide]


An elephant and a lion are currently 1 mile apart. The elephant runs directly away from the lion at 19 miles per hour, while the lion runs directly towards the elephant at 24 miles per hour.  How many minutes will it take for the lion to catch the elephant?

Every hour the lion gains $ 24\minus{}19\equal{}5$ miles on the elephant.  It only needs to gain $ 1$ mile, so it will take $ \frac{60}{5}\equal{}\boxed{12}$ minutes.


no one answers it? Is that because i did not translate correctly and clearly?

I am dumb.  what is constant power?  just constant work in a given time interval or constant acceleration by distance?

[quote="warut_suk"]Is it correct, shobber?[/quote]
Can you show your steps? I don't think it is correct...

Well, I wasn't sure when I read the problem. How is $a$ related? It's from velocity $0$ to $v_1$, not $v_1$ to $v_2$. So... can you show how to do it?

you assumed constant acceleration, not constant power. Are they the same?

[quote="warut_suk"]By the way, my work is:
$s = v_1t+\frac{1}{2}at^2$, but since $a = \frac{v_2-v_1}{t}$, we have $s = v_1t+\frac{1}{2}(v_2-v_1)t = \frac{1}{2}(v_1+v_2)t$[/quote]
well, $0 \to v_1$ with constant acceleration $a$, but $v_1 \to v_2$ is only with constant power, not constant acceleration.

while at first I thought that the first part, where the car accelerates from 0 to $v_1$ was not important, but after running through some math it turns out that the initial acceleration at $v_1$ appears to come into play when calculating the time from $v_1$ to $v_2$.

[quote="shobber"]A car starts from rest with constant acceleration $a$, when it gets to velocity $v_1$, it will move with constant power. And it stops accerlerating when the velocity becomes $v_2$. The displacement of the car when it accelerates from $v_1$ to $v_2$ is $s$. Find the time $t$ the car takes to accelerate from $v_1$ to $v_2$.[/quote]

[hide]

Since we want the time between $v_1$ and $v_2$ what happens before the car reaches $v_1$ is irrelevant. That the car moves with constant power between $v_1$ and $v_2$ implies that the acceleration between $v_1$ and $v_2$ is not constant.  However, we can write down by inspection:

$v_2=v_1+\int a(t) \mathrm{d}t \\
=v_1+\langle a \rangle \Delta t$

We also know that $v_2^2 = v_1^2 + 2 \langle a \rangle s$, so that

$\langle a \rangle = \frac{v_2^2-v_1^2}{2s}$

$\Rightarrow \boxed{ \Delta t = \frac{2s}{v_2+v_1}}$

[/hide]

well, i don't know how to solve it with calculus. But we can use energy conservation to do it. I shall post my solution later.

[quote="shobber"][quote="warut_suk"]Then I'm not sure what "constant power" is...[/quote]
$P=Fv$, if $P$ is constant and $v$ increases, then $F$ decreases, so $a$ decreases.[/quote]
I think I haven't learnt it yet.  :(

[quote="warut_suk"][quote="Spoon"][quote="Dr. No"]

We also know that $v_2^2 = v_1^2 + 2 \langle a \rangle s$, so that

[/quote]

Could you please show this? For instance, in the case $a(t) = at + a_0$, I don't see how it holds, where $\langle a \rangle = \frac{at}{2} + a_0$.[/quote]

Isn't the well-known formula $v_2^2 = v_1^2 + 2as$?[/quote]

First, the well known formula is $v_2^2 = v_1^2 + 2as$ , which follows from the change in kinetic energy equals to the work done, but in the form given it is only valid for constant acceleration. We don't have constant acceleration over the interval $s$. The power is constant:

$\text{Power} = \vec{F}\cdot \vec{v}$.

As $\vec{v}$ increases $\vec{F}$ decreases, and since $\vec{F} \propto \vec{a}$ then $\vec{a}$ decreases as well.

However, the Mean Value Theorem gives the result that $\langle a \rangle \Delta t = \int_t^{t+\Delta t} a(t) \mathrm{d}t$.

The mean value, $\langle a \rangle$, is equal to the value of $a(t)$ if $a(t)$ were constant. Therefore, I can substitute it for $a$ in the equation $v_2^2 = v_1^2 + 2as$.

When I do that the result is $\Delta t = \frac{2s}{v_2+v_1}$ regardless of the shape of the function $a(t)$.

[quote="warut_suk"]Isn't the well-known formula $v_2^2 = v_1^2 + 2as$?[/quote]

Yes, but that assumes a constant acceleration, and it is easy for anyone to show. I was not able to "easily" show anything related to that when instead made the system have a constant jerk. By $\langle a \rangle$ I assume Dr. No means average value.

I agree with $v_1 = v_0 +  \langle a \rangle t_1$ as it follows directly from it's definition, but the above equation is not as simple, at least from what I can see.

Edit: I had not realized there was a page 2 when I posted this.

[quote="Spoon"]
I ask you again to consider $a(t) = at+a_0$ as I cannot derive an equation that looks like: $v_1^2 = v_0^2 + 2(\frac{at}{2}+a_0)s$[/quote]

Sigh, look. Take $a(t) = at+a_0$ and derive $v_1^2 = v_0^2 + 2(\frac{at}{2}+a_0)s$ without using any assumptions. That is all I am asking you to show. Stop giving the answer to Shobber's problem, as it's not relavent to my question and I obviously know by now that you think it's the solution. I am asking you this because I would really like to see it because I cannot figure out how to produce it.

My question assumes that at $t=0, v = v_0$ as I'm sure is clear to you.

Let's assume $P = kt$, let's see what we get.

We have $P = Fv$ which is equal to $P = \frac{m}{2}\frac{d}{dt}(v^2)$

so we have

$\frac{m}{2}\frac{d}{dt}(v^2) = kt$
$\frac{d}{dt}(v^2) = \frac{2kt}{m}$
or
$v^2 = \frac{kt^2}{m}$
or
$v = \sqrt{\frac{k}{m}}t$
and you'll notice
$a = \sqrt{\frac{k}{m}}$ <- should look familiar to most people. Basically we started with power and we've "derived" the equation of motion. Then we easily go on to show a number of things such as th equation under debate.

Now assume $P=k$

If we apply the exact same logic, we get:

$v = \sqrt{\frac{2kt}{m}}$
$a = \frac{k}{m}{\left(\frac{2kt}{m}\right)}^{-1/2}$

You will notice that $a$ decreases with time, as it should. These are the equations of motion for an object traveling with constant power.

If you use these equations along with the position equation (not shown) to calculate the time it takes, you will get:

$s = \frac{2(v_1-v_0)}{3\sqrt{t}}+v_0t$

which cannot be solved.

[quote="Spoon"]You have merely shown me that if $a$ in the "well known equation" is the mean (average) value then you can substitute in the mean value...

You neglect that we prove the "well known equation" with the premiss that accleration is constant. Why would it hold in general? I ask you again to consider $a(t) = at+a_0$ as I cannot derive an equation that looks like: $v_1^2 = v_0^2 + 2(\frac{at}{2}+a_0)s$[/quote]

First, I must say that your notation is confusing. You use $a$ to denote the acceleration as a function of time, and its proportionality to time (in your example) as well. Be that as it may,

$ Let \, a(t) = at+a_0$

$\langle a \rangle =\frac {\int_t^{t+\Delta t} a(t')\mathrm{d}t'}{\Delta t} \\
= \frac{\frac{1}{2}a[(t+\Delta t)^2-t^2]+a_0 \Delta t}{\Delta t} \\
= at +a_0$

According to how the limits of integration were chosen, $t$ is the specific moment at which the speed of the car is $v_1$, therefore $\langle a \rangle$ is constant. Since $\langle a \rangle$ is constant the "well known equation" is valid and it follows that $\Delta t = \frac{2s}{v_2+v_1}$.

[quote="Spoon"][quote="Dr. No"]
$\langle a \rangle =\frac {\int_t^{t+\Delta t} a(t')\mathrm{d}t'}{\Delta t} \\
= \frac{\frac{1}{2}a[(t+\Delta t)^2-t^2]+a_0 \Delta t}{\Delta t} \\
= at +a_0$
[/quote]

Recheck your math, this does not simply to $at +a_0$

It's

$at +\frac{a\Delta t}{2}+a_0$[/quote]

Indeed.  :oops:  However,  $at +\frac{a\Delta t}{2}+a_0$ may be recast as $\frac{1}{2}[a(t +\Delta t) + a_0]+\frac{1}{2}[at+a_0]$, which is equal to the arithmetic mean of the accelerations at the end points of the interval and is still constant. The result $\Delta t =\frac{2s}{v_2+v_1}$ still holds.

[quote="Dr. No"][quote="Spoon"]Let's assume $P = kt$, let's see what we get. [/quote]

You go wrong at your very first step. Your assumption implies that $t$ is proportional to $\vec{a}\cdot \vec{v}$.  There is nothing in the problem that states this. Even if it were the case, the problem can still be solved. $P = kt$ results in a separable 1st order ODE the only thing not known in solution of which is the proportionality constant that you introduced.[/quote]

Shobber asked the question implying 1 dimension.

[quote="Spoon"][quote="Spoon"]
I ask you again to consider $a(t) = at+a_0$ as I cannot derive an equation that looks like: $v_1^2 = v_0^2 + 2(\frac{at}{2}+a_0)s$[/quote]

Sigh, look. Take $a(t) = a_1t+a_0$ and derive $v_1^2 = v_0^2 + 2\left(\frac{a_1t}{2}+a_0\right)s$ without using any assumptions. That is all I am asking you to show. Stop giving the answer to Shobber's problem, as it's not relavent to my question and I obviously know by now that you think it's the solution. I am asking you this because I would really like to see it because I cannot figure out how to produce it.

My question assumes that at $t=0, v = v_0$ as I'm sure is clear to you.[/quote]

Let's start from first principles.

Work, $W = \int \vec{F} \cdot \mathrm{d} \vec{s}$

Power, $P = \frac{\mathrm{d}W}{\mathrm{d}t}$

From Leibniz's Integral formula : $P = \frac{dW}{dt} =\frac{d \vec{s}}{dt} \cdot \nabla \int \vec{F} \cdot d \vec{s} = \vec{v} \cdot \vec{F}$

$\vec{v} \cdot \vec{F} = \vec{v} \cdot m \frac{d \vec{v}}{dt}$

$W = \int_{t_1}^{t_2}m \vec{v} \cdot \frac{d \vec{v}}{dt} dt = \int_{s_1}^{s_2} \vec{F}(\vec{s}) \cdot d \vec{s}$

Changing the limits of integration on the left hand integral and dividing through by $m$ we obtain:

$\int_{v_1}^{v_2} \vec{v} \cdot d \vec{v} = \int_{s_1}^{s_2} \vec{a}(\vec{s}) \cdot d \vec{s}$, which implies:

$v_2^2 = v_1^2 +2 \int_{s_1}^{s_2} \vec{a}(\vec{s}) \cdot d \vec{s}$

The integral, $\int_{s_1}^{s_2} \vec{a}(\vec{s}) \cdot d \vec{s} = \langle a \rangle \Delta s$ by the Mean Value Theorem.

I contend that whether one integrates over $t$ or whether one integrates over $s$ makes no difference; $\langle a \rangle$ is the same in either case.

ok guys,  i don't know much about calculus, but we can use the basic things to solve this problem.
Assume $f$ the friction of all time, then:
$P=f v_2$
$P=(f+ma)v_1$
$Pt-fs=\frac{1}{2} m (v_2^2-v_1^2)$

Solve the system and cancel $m$, we get the result.

Dr. No. I see now. Thanks, I will agree with your answer now.

Shobber: You should explain stuff that you put down rather than saying "You have these equations and you solve for blah"

And also, you should solve them and post your solution :).

OK...
From the definition of power, we have $P=Fv$. But since $F-f=ma$, so $P=(f+ma)v_1$ is the power of the car at velocity $v_1$.

Then it will move with this constant power. So when it stops accelerating, we have $F=f-ma=f$, so $P$ has another expression $P=fv_2$.

Also, according to the conservation of kinetic energy, we know $Pt-fs=\frac{1}{2}m (v_2^2-v_1^2)$
Then we can choose to cancel $m$ or $f$. I canceled $f$ and get:
\[v_2t=\frac{v_2-v_1}{2av_1}(v_2^2-v_1^2)+s\]

thank you :)

another note... you always say you don't solve anything with calculus... but in reality, all you do is take the equations derived from calculus as "known" and then manipulate them. When someone uses calculus, they are doing no more than you, except generally more "advanced" manipulation COULD be performed (though rarely needed).

You immediately state "P=Fv", but you couldn't know that if not derived using calculus. There are a few basic exceptions that can be done purely on algebraic manipulation, but it's ugly and not robust imo.

Well, i cannot understand the calculus stuff you guys posted since i know few about calculus. I only know to use integral to find sth, but i never learned any advanced definitions of physics...We do not study them in Chinese high school.
And for $P=Fv$, my book goes like that: $P=\frac{W}{t}=\frac{Fs}{v}$, if $s, v \to 0$, then we have instant power $P=F\frac{ds}{dt}=Fv$.

It seems I shall have time to learn calculus application in physics this summer vacation. So do you know any website or e-books from which I can learn?

[quote="shobber"]Well, i cannot understand the calculus stuff you guys posted since i know few about calculus. I only know to use integral to find sth, but i never learned any advanced definitions of physics...We do not study them in Chinese high school.[/quote]

shobber,

It might behoove you to study some calculus. Physics, especially mechanics, is largely a course in applied calculus. Historically speaking, the calculus exists because Issac Newton lacked the mathematics to solve experimental problems he encountered in his studies of motion. Starting with no more mathematical background than highschool arithmetic, he taught himself algebra and subsequently invented the calculus.

As an aside, Richard Feynmann spent some time as a visting lecturer in Brazil. His chief criticism of Brazilian students he met was that a number of them had memorized physics formulas but seemed to have no intuitive understanding of why the formulas worked.

To that I will only add if you're going to be a physicist, be an excellent one. The world is too full of people guided by mediocre lights.

[quote="Spoon"][quote="Dr. No"]As an aside, Richard Feynmann spent some time as a visting lecturer in Brazil. His chief criticism of Brazilian students he met was that a number of them had memorized physics formulas but seemed to have no intuitive understanding of why the formulas worked.[/quote]

This was probably one of the most entertaining parts of "Surely you're Joking Mr. Feynmann."[/quote]

I thought the funniest part was his account of the psychiatric interview in his medical exam for the Draft Board.  :lol:

lol, oh that was funny indeed.

Of course we can't forget when he was looking at the blueprints and had no idea what he was looking at, pointed to something he thought might be a window and found a flaw in the plans. Now why can't stuff like that happen to me.


Prove that in a triangle $ ABC$, $ \sin A \sin B \sin C \leq \frac{3\sqrt{3}}{8}$.

If any of $ A,B,C$ is 0, then we are done, since we have $ 0 < RHS$.
Otherwise, $ f(x) \equal{} \ln \sin x$ is concave over $ (0,\pi)$, so $ \ln \sin A \plus{} \ln \sin B \plus{} \ln \sin C \le 3 \ln \sin \frac{A\plus{}B\plus{}C}{3} \equal{} 3 \ln \frac{\sqrt{3}}{2}$. Raising both sides to the power of $ e$, we obtain the desired result.

hello, by the AM-GM inequality and the Jensen inequality we get
$ {\sqrt[3]{\sin(A)\sin(B)\sin(C)}\le \frac{\sin(A)+\sin(B)+\sin(C)}{3}\le\sin\left(\frac{A+B+C}{3}\right)=\sin\left(\frac{\pi}{3}\right)= \frac{\sqrt{3}}{2}}$
from here you will get your theorem.
Sonnhard.


Prove that for all $ a,b,c$ be nonnegative real numbers, we have
\[ \frac{a^2}{b\plus{}c}\plus{}\frac{b^2}{c\plus{}a}\plus{}\frac{c^2}{a\plus{}b}\ge\frac2{3}\frac{(a\plus{}b\plus{}c)^2}{ab\plus{}bc\plus{}ca}\left(\frac{a^2}{2a\plus{}b\plus{}c}\plus{}\frac{b^2}{2b\plus{}c\plus{}a}\plus{}\frac{c^2}{2c\plus{}a\plus{}b}\right)
\]

[hide="@manlio"]Dear Manlio,

I would like to dedicate this Inequality to you for your interest to other Inequalities of mine. Hope you enjoy it. :)[/hide]

General problem for you,

Prove that for all $ a,b,c$ be nonnegative real numbers and $ k\ge2$, we have
\[ \frac {a^k}{b \plus{} c} \plus{} \frac {b^k}{c \plus{} a} \plus{} \frac {c^k}{a \plus{} b}\ge\frac2{3}\frac {(a \plus{} b \plus{} c)^2}{ab \plus{} bc \plus{} ca}\left(\frac {a^k}{2a \plus{} b \plus{} c} \plus{} \frac {b^k}{2b \plus{} c \plus{} a} \plus{} \frac {c^k}{2c \plus{} a \plus{} b}\right)
\]


function
integration
factorial
probability
calculus
real analysis
derivative

Which is the inverse Laplace of  [1/(s+1)]^N?

thx

The Laplace transform of $t^{n}$ is $\frac{n!}{s^{n+1}}.$ More generally, the Laplace transform of $t^{\alpha}$ (assuming $\alpha>-1$) is $\frac{\Gamma(\alpha+1)}{s^{\alpha+1}}.$

Shifting the Laplace transform - that is, replacing $s$ everywhere by $s+1$ - is the result of multiplying the original function by an exponential.

Put all that together and we find that the inverse Laplace transform of

$\frac{1}{(s+1)^{N}}$ is $\frac{1}{(N-1)!}t^{N-1}e^{-t}.$

what is the function Γ? and where i am replacing the s with s+1?

Could you be more specific please?

thx!

The Gamma function:

For $\alpha>0,\ \Gamma(\alpha)=\int_{0}^{\infty}x^{\alpha-1}e^{-x}\,dx.$

Integrate by parts, and you get the all-important recursion formula:

$\Gamma(\alpha+1)=\alpha\Gamma(\alpha).$

From this and from the fact that $\Gamma(1)=1,$ we find that $\Gamma(n+1)=n!.$ So we can consider this as a continuous interpolation of the factorial function.

It's also possible to calculate that $\Gamma\left(\frac12\right)= \sqrt{\pi}.$ Hence:

$\Gamma\left(\frac32\right)=\frac{\sqrt{\pi}}2, \Gamma\left(\frac52\right)=\frac32\cdot \frac12\cdot\sqrt{\pi},$ and so on.

A user of applied mathematics (differential equations, probability, statistics, etc.) should have a working familiarity with the Gamma function. It's too widespread and too useful to ignore.

(Yes, the Gamma function can be extended as an analytic function to the whole complex plane except for the poles at the nonpositive integers. Leave that lesson for another time.)

--

As for the second part, I was referring to this theorem:

If $f(t)$ has Laplace transform $F(s)$ and if $g(t)=e^{at}f(t),$ then the Laplace tranform of $g$ is $G(s)=F(s-a).$

The proof of this is an immediate consequence of the integal defnition of the Laplace transform. Just write it all out and look at what you have.

Work out for yourself how I was applying this theorem. Look carefully and what we have and think about it.

And how can i calculate the pdf of the sum ax+by ,a,b deterministic constants,and x,y random variables with exponential distribution,with mx and my the rates?

thx

You didn't say anything about the relationship of $X$ and $Y.$ We need something, or we won't be able to solve the problem. I'll make the simplest assumtion, namely that $X$ and $Y$ are independent.

We can do what you want directly, using a convolution integral, or we can use a transform method. For a transform method, I'll use moment-generating functions. These are actually Laplace transforms, but the sign conventions and notations differ somewhat from the way Laplace transforms are used in differential equations.

For a random variable $X,$ define its moment generating function as $m_{X}(t)=E(e^{tX}).$ It's called the moment generating function because $m(0)=1,\ m'(0)=E(X),\ m''(0)=E(X^{2})$ and in general, $m^{[k]}(0)=E(X^{k}).$ (That's assuming that these derivatives and moments exist.)

If $U=aX$ then $m_{U}(t)=E(e^{t(aX)})=E(e^{(at)X})=m_{X}(at).$

If $X$ and $Y$ are independent and $U=X+Y$ then

$m_{U}(t)=E(e^{t(X+Y)})=E(e^{tX})E(e^{tY})=m_{X}(t)m_{Y}(t).$

If $X$ is exponentially distributed with mean $\mu,$ then its pdf is $\frac1{\mu}e^{-x/\mu}$ for $x>0$ and zero otherwise. Its moment-generating function is

$m_{X}(t)=\int_{0}^{\infty}e^{tx}e^{-x/\mu}\,dx=\frac1{1-\mu t}.$

Now we have $X$ and $Y$ independent exponential distributions with means $\mu_{X}$ and $\mu_{Y}.$ Let $U=aX+bY.$

$m_{U}(X)=\frac{1}{(1-a\mu_{X}t)(1-b\mu_{Y}t)}.$

From here, there are two cases.

Case 1: $a\mu_{X}\ne b\mu_{Y}.$

By partial fractions,

$\frac{1}{(1-a\mu_{X}t)(1-b\mu_{Y}t)}=-\frac{a\mu_{X}}{b\mu_{Y}-a\mu_{X}}\cdot\frac{1}{1-a\mu_{X}t}+\frac{b\mu_{Y}}{b\mu_{Y}-a\mu_{X}}\cdot\frac{1}{1-b\mu_{Y}t}$

This makes $U$ a linear combination of exponential distributions. Using the formula derived above, only backward, we get the following pdf for $U:$

$f_{U}(u)=\frac{1}{b\mu_{Y}-a\mu_{X}}\left(-e^{-u/(a\mu_{x})}+e^{-u/(b\mu_{Y})}\right)$ for $u>0$ and zero otherwise.

Case 2: if $a\mu_{X}=b\mu_{Y},$ then we get a version of a gamma distribution:

$f_{U}(u)=\frac{1}{(a\mu_{X})^{2}}ue^{-u/(a\mu_{X})}.$

i forgot to say that are indeed independent,i see,you are right...i expanded it for N random variables and is correct

But what if.. suppose we have a Gaussian rv with mean=1 and variance=1.We create N samples χ1,...χΝ with this distribution and we try to calculate the average (that is 1) by using the arithmetic average
μ=(χ1+...+χΝ)/Ν.
α)μ is a rv.What is the mean and the variance of μ?

β)Is the result of the mean expected?

c)What can we say about the variance?What should be the variance in order to say that we have an excellent estimation?

d)and what is the N in order to have one decimal digit accurance at the calculation of the mean?

i want some help ...

thx!


Hi to all of you, I'm new here...
Can somebody help me how to fill in these numbers to this chess board?
I attach the file below, name ChessBoard.xls.
Thanks.

Shouldn't this be moved into Math and Strategy Games section?
Dini, I think you would have more help there...a bunch of chess geniuses there. :wink:


geometry
3D geometry
trigonometry
vector

There is a cubic block having mass $ m$ that is sliding along the sides of a right angle wedge where the angle between the wedge and the horizontal ground is $ \theta$. If the coefficient of kinetic friction between all contacting surfaces of the block and the surface of the wedge is $ \mu$, find the sliding acceleration of the block.

My attempts:
1)I thought that the friction against the motion of this cube would go backwards (against the sliding of the block, which would be $ mg\sin \theta$). My original approach, therefore was this.
$ F \equal{} ma$
$ F_s \minus{} 2 Fr \equal{} ma$
$ mg\sin\theta \minus{} 2mg\cos\theta\mu \equal{} ma$
$ a \equal{} g(\sin\theta \minus{} 2 \mu\cos\theta)$

2) But then I thought that the normal forces to the surfaces would be perpendicular too - since the two sides touching the wedge are themselves perpendicular. So now i tried this with two friction forces that are acting in different angles - one at a perpendicular to the other. My new calculations:
$ mg\sin\theta \minus{} mg\mu\sin\theta \minus{} mg\mu\cos\theta \equal{} ma$
$ a \equal{} g\sin\theta \minus{} g\mu\sin\theta \minus{} g\mu\cos\theta$

I'm not sure if any of these is right: i'm not sure how to deal with two friction forces acting on a 3-dimensional body. I'd really appreciate any help. Thanks  :)

You are right that both friction forces will be opposing the direction of motion of the cube.  So, even though the normal forces are perpendicular, the friction forces that arise from them will be parallel.  The difficulty will be calculating the normal forces.  I recommend the following approach: imagine there was just a ramp, not a wedge.  What would the normal force be?  The two normal forces we have in this problem have to add up to the normal force in the situation with just a ramp.

Thanks, inteluser.

I'm still not sure how to calculate the normal forces though.
If we call the first normal force $ N_1$ and the second $ N_2$, then:
$ N_1 + N_2 = N{}_t{}_o{}_t{}_a{}_l = mg\mu_k\cos\theta$
Can you give me a hint about how to calculate $ N_1$ and $ N_2$?

Also, is calculating these forces necessary to find the total friction?
Wouldn't the value of the total $ fr$ be $ N_1\mu_k + N_2\mu_k = (N_1 + N_2)(\mu_k) = N{}_t{}_o{}_t{}_a{}_l\mu_k$? - the same as if we had just considered the block on a ramp idea?

This seems intuitive since both frictions are pointing the same way, and the two normal forces are equal to the normal force of the block as if it were on a ramp.

[quote="xsk13"]I'm still not sure how to calculate the normal forces though.
If we call the first normal force $ N_1$ and the second $ N_2$, then:
$ N_1 + N_2 = N{}_t{}_o{}_t{}_a{}_l = mg\mu_k\cos\theta$
Can you give me a hint about how to calculate $ N_1$ and $ N_2$?[/quote]
I expect you meant $ mg\cos\theta$ rather than $ mg\mu_k\cos\theta$?  In that case,
we have $ \vec N_1+\vec N_2=\vec N_{\text{total}}$, but this is vector addition.  It
may help to note that $ \vec N_1$ is at a $ 45^\circ$ angle to $ \vec N_{\text{total}}$ by symmetry.

[quote="xsk13"]Also, is calculating these forces necessary to find the total friction?
Wouldn't the value of the total $ fr$ be $ N_1\mu_k + N_2\mu_k = (N_1 + N_2)(\mu_k) = N{}_t{}_o{}_t{}_a{}_l\mu_k$? - the same as if we had just considered the block on a ramp idea? [/quote]

It's true that $ \left|\vec f_r\right|=\left(\left|\vec N_1\right|+\left|\vec N_2\right|\right)\mu_k$, but $ \left|\vec N_1\right|+\left|\vec N_2\right|\ne\left|\vec N_{\text{total}}\right|$, even though $ \vec N_1+\vec N_2=\vec N_{\text{total}}$.

[quote="inteluser"][quote="xsk13"]I'm still not sure how to calculate the normal forces though.
If we call the first normal force $ N_1$ and the second $ N_2$, then:
$ N_1 + N_2 = N{}_t{}_o{}_t{}_a{}_l = mg\mu_k\cos\theta$
Can you give me a hint about how to calculate $ N_1$ and $ N_2$?[/quote]
I expect you meant $ mg\cos\theta$ rather than $ mg\mu_k\cos\theta$?  In that case,
we have $ \vec N_1 + \vec N_2 = \vec N_{\text{total}}$, but this is vector addition.  It
may help to note that $ \vec N_1$ is at a $ 45^\circ$ angle to $ \vec N_{\text{total}}$ by symmetry.[/quote]

Yeah, sorry. I meant $ mg\cos\theta$. Ahh I see.

[quote="inteluser"][quote="xsk13"]Also, is calculating these forces necessary to find the total friction?
Wouldn't the value of the total $ fr$ be $ N_1\mu_k + N_2\mu_k = (N_1 + N_2)(\mu_k) = N{}_t{}_o{}_t{}_a{}_l\mu_k$? - the same as if we had just considered the block on a ramp idea? [/quote]

It's true that $ \left|\vec f_r\right| = \left(\left|\vec N_1\right| + \left|\vec N_2\right|\right)\mu_k$, but $ \left|\vec N_1\right| + \left|\vec N_2\right|\ne\left|\vec N_{\text{total}}\right|$, even though $ \vec N_1 + \vec N_2 = \vec N_{\text{total}}$.[/quote]
Thanks so much. I forgot to factor in the difference in directions of the normal forces - and that they are vectors. I'll try to redo this with that info in mind.

Edit: My new calculations:
$ \vec N_1 + \vec N_2 = \vec N_{\text{total}}$
$ \sqrt2 \vec N_1 = \vec N_{\text{total}}$

$ mg\sin\theta - N_{\text{total}} = ma$
$ mg\sin\theta - \sqrt2mg\mu_k\cos\theta = ma$
$ a = g(\sin\theta - \sqrt2\mu_k\cos\theta)$


Prove the following :

1.  a and b are real numbers .If 0<b<a then a + 1/(a-b)b >= 3.

2.a is real and >0.Then ( a^4 +9)/10a >4/5

3.a and b are positive real numbers .Then 3(a^3) + 7(b^3) >9(a)(b^2)

These inequalities are proved by the authors using AM-GM inequality.

3 can be strengthened to $3a^3 + 6b^3 \geq 9ab^2$

[quote="kueh"]2.
[hide] ${a^4 + 1} \geq 2a^2$
Thus, $\frac{a^4 + 10}{10a} \geq \frac{2a^2 + 8 - 8a + 8a}{10a} = \frac{2(a-2)^2}{10a} + \frac{4}{5} \geq \frac{4}{5}$

But we note that the last inequality holds when a =2 and the first when a=1 only. Thus equality cannot hold it both case.[/hide]

What is the best bound for question 2?[/quote]
$2\frac{3^\frac{3}{4}}{5}$
and I think you can slove it yourself :)


quadratics
MATHCOUNTS
algebra
quadratic formula

Do you need to know how to find the roots of quadratic equations for mathcounts?

I don't remember needing to know this, but I was looking through some old competitions (from 15 years ago) and this knoledge was needed then.

Has this changed?  Is there an official list somewhere of what topics mathcounts covers?

Thanks!

You don't need to use quadratic formula for mathcounts problems but sometimes it's a quicker way of doing the problem.  Although, you do need to know how to factor equations.

Yes, I remember several instances while taking Mathcounts tests and I had to solve quadratic equations.  They were all whole numbers I believe, mostly solved with factoring, although as the above post says, quadratic formula is not needed but sometimes is faster.

The quadratic formula is:

[-b +/- sqrt(b^2 - 4ac)]/2a
where
ax^2 + bx + c = 0.

Of course, if you can factor it, it'll be a lot easier than plugging those numbers in :P .

If b^2 - 4ac is negative, there are no solutions.

i believe they have taken problems like that out of competitions. i recall coming across a couple from about 14 years ago that required you to know what the :rt2: was out to 4-5 digits. i don't think they'll require you to do anything like that now. like soilent green said, it can sometimes be much easier than factoring out the problem!

You don't have to do it but if you have time, why CAN't YOU?

And by the way, here is the derivation of the quadratic formula.

\[ax^2+bx+c = 0\]
\[x^2+ \frac {b}{a}x + \frac {c}{a} = 0\]
\[x^2+ \frac {b}{a}x = - \frac {c}{a}\]
\[x^2+ \frac {b}{a}x + \frac {b^2}{4a^2} = \frac {b^2-4ac}{4a^2}\]
\[{(x+ \frac {b}{2a})}^2 = \frac {b^2-4ac}{4a^2}\]
\[x+ \frac {b}{2a} =\frac   {\pm \sqrt {b^2-4ac}}{2a}\]
\[x = \frac {-b \pm {\sqrt {b^2-4ac}}}{2a}\]


What happens if two people got exactly the same questions right. What does Mathcounts do?

I think it depends on the official, but it might depend on:

Who was better in the countdown
Who got the first question wrong on the sprint
Who got the last question right on the sprint

Course, the last two only apply if they got the same score, but not the same questions right

What is the probabilty of two people getting the exact same problems right?  There are 30 sprint, 8 target, and let's say team doesn't count (by the way, in response to your question, maybe they would look at team scores).  Neither does countdown.  Figure it out!

If they're not exactly the same, they start from the bottom of the sprint round since they get harder at the end. The first person to miss one loses the tie.

If they happened to be the same...which is highly unlikely....and if it mattered (such as getting top 4 at state to go to nats. or top 2 at chapter to go to state, or top 3 at nats. to get a prize) then they would be put head-to-head during the countdown round.

There were two 46's at the Houston Chapter. Magicmaniax told me the way they did it is that they just gave them more problems, and I would guess it's the person who gets more of them right...

Yea...
In our chapter, they just gave us two questions on a sheet of paper. We tried to finish them as fast as we could. Whoever got the most right won, however, if there was a tie, then the person who finished first won. If no one any of the two questions right, then we went on to a second set of two questions, but ours ended fairly quickly.

You can read about the tiebreaking rules on the MATHCOUNTS Web Site in the [url=http://www.mathcounts.org/webarticles/articlefiles/510-C&SGenInstruct.pdf]General Instructions[/url] document.

In my six years of coaching, neither my Chapter nor State has ever used the Tiebreaker Round.  They have reported ties (for example there was a tie for 132nd place).  But they only use the Tiebreaker Round when ranking matters for Countdown position or a prize.  What Magixmaniac said about the Tiebreaker Round is more than I knew.

Andrew Chien and Eric Chen had perfect scores in CA in 2003.

Here's a link to how special tie breakers were used in 2003:

http://www.cspescvc.org/newsletters/CSPE06_Spring_2003_NL.PDF

Naga

[quote="dwx314"]I think it goes like so:
1: Who got most sprints
2: Correct 21-30
3: Backwards from 30-21
That's all I know, I think it conitnues like that to 1, and then goes to targets, and then the tiebreaker round, but I'm probably wrong[/quote]
Your first three rules match what is given in the [url=http://www.mathcounts.org/webarticles/articlefiles/510-C&SGenInstruct.pdf]General Instructions[/url].  The Target questions are the 4th priority.  Just click the link, and go to page 9 of 12 to get the definitive tiebreaking rules.

Which, for those who have slow connections/no pdf/etc., this is what they say:
1.Sprint Score
2.Total correct form probs 30-21
3.Correct in order from 30-21
4.Target Rounds Compared in order 8-1
5.Total correct in 20-11
6 Correct in order from 20-11
7.Correct in order from 10-1
After that, it off to the tie-breaker rounds.


Let ABCD be a cyclic quadilateral, O be the intersection of its diagonals AC and BD. Let M, N, P, Q be projections of O respectively on AB, BC,CD,DA. Prove that
     MP is perpendicular to NQ $\Longleftrightarrow$ ABCD is a trapezoid.

We use to easy-to-prove facts:

(1) $MNPQ$ is a circumscribed quadrilateral.
(2) A circumscirbed quadrilateral's diagonals are perpendicular iff one of them is a symmetry axis for the quadrilateral.

After combining these we get the solution.


The first inequality is easy.
We have 2bc(=1/2(b+c))*2 so inequality is equivalent to
(a+1)/(b+c)*2 + (b+1)/(a+c)*2 + (c+1)/(a+b)*2)=9
Last inequality can be easily proved by Jensen ineq.since funct. (a+1)/(1-a)*2 (=a+1/(b+c)*2 becouse a+b+c=1) is convex for all a from interval(0,1)
Equality holds when a=b=c=1/3


Happy Easter!

What really does Easter celebrate?


limit
real analysis
real analysis unsolved

Let F be a field with the least upper bound property. Prove that every Cauchy sequence in F converges in F.

Let's do the case $\mathbb{F}=\mathbb{R}$. (I am not enough of an algebraist to handle the general case). For all $\varepsilon > 0$, there is an $N$ such
that for all $m, n \geq N$ \begin{eqnarray*}|x_{m}-x_{n}| & < & \varepsilon \end{eqnarray*} Clearly $\left( x_{n}\right)$ is upper bounded. If not, then for all $M \in \mathbb{R}$ there is an $a_{j}$ such that $a_{j}\geq M$.
In particular pick $\varepsilon$ and an $N$ such that for all $m \geq N$ we have \begin{eqnarray*}| x_{m}-x_{N}| & < & \varepsilon \end{eqnarray*} Pick $m \geq N$ such that $x_{m}> \varepsilon+x_{N}$ (exist since $x_{m}$ unbounded). Then $\varepsilon < | x_{m}-x_{N}| < \varepsilon$. A contradiction.
WLOG suppose that there is an infinite increasing subsequence $\left( x_{i_{j}}\right)_{j = 1}^{\infty}$ of $\left( x_{j}\right)$. Then \begin{eqnarray*}S & = & \left\{ x_{i_{j}}: j \in \mathbb{N}\right\}\end{eqnarray*} is upper bounded. Therefore $\lim x_{i_{j}}= \sup S$ (proved this in another topic you posted). Hence for all $\varepsilon > 0$, there is an $N$ such that for all $m, i_{j}\geq N$ we have \begin{eqnarray*}| x_{m}-x_{i_{j}}| & < & \varepsilon \end{eqnarray*} Observe that \begin{eqnarray*}| x_{m}-x_{i_{j}}| & < & \varepsilon\\ | x_{m}-x_{i_{j+1}}| & < & \varepsilon\\ \ldots & & \\ \lim_{j \rightarrow \infty}| x_{m}-x_{i_{j}}| = | x_{m}-\sup S | & < & \varepsilon \end{eqnarray*} Therefore for all $\varepsilon > 0$, there is an $N$ such that for all $m \geq N$ we have \begin{eqnarray*}|x_{m}-\sup S | & < & \varepsilon \end{eqnarray*} Hence $\lim x_{n}= \sup S$. The Least Upper Bound axiom, is sometimes called the Completeness axiom, and this is somewhat more suggestive for this problem. I hope the generalization to an arbitrary field $\mathbb{F}$ should not be a problem ... :)


geometry
trigonometry
symmetry
perpendicular bisector
geometry unsolved

i tried solving this but..i can't. hope you can help me with it.

you are supposed to solve for [b][color=red]a[/color][/b] by using [b]elementary geometry [/b](i.e. 180 degrees rule, etc.) and not by [b]trigonometrical laws[/b] (sine or cosine laws).

you can visit [i]http://thinkzone.wlonk.com/MathFun/Triangle1Hint0.htm[/i] for the scaled drawing.

The usual recipe is to draw more lines to create more isosceles triangles, until an equilateral triangle is found, which allows to find even more isosceles triangles.  First, $\angle ABC = \angle BAC = 80^\circ,$ $\angle ACB = 180^\circ - (\angle ABC + \angle BAC) = 20^\circ,$ and the triangle $\triangle ABC$ is isosceles with AC = BC.  Let the perpendicular bisector of the segment AB, identical with the bisector of the angle $\angle ACB,$ meet the lines AE, BD at points F, G, respectively and let a parallel to the line AB through the point D meet the line BC at a point D'. Because of symmetry, the lines AD', BD meet at the point G on the perpendicular bisector of the segment AB. The triangles $\triangle AFC,\ \triangle AD'C$ are both isosceles, because

$\angle CAF \equiv \angle CAE = 10^\circ,\ \ \angle ACF = \frac{\angle ACB}{2} = 10^\circ$

$\angle ACD' \equiv \angle ACB = 20^\circ,\ \ \angle CAD' = \angle CBD = \angle CBA - \angle ABD = 20^\circ$

Hence, AF = CF and AD' = CD'. The triangles $\triangle AFG \cong \triangle CFE$ are congruent by ASA, because AF = CF, $\angle AFG = \angle CFE$ and $\angle FAG = \angle FCE = 10^\circ$. Consequently, AG = CE and $GD' = AD' - AG = CD' - CE = ED'.$ The triangles $\triangle ABG,\ \triangle DD'G$ are both equilateral (obviously). Therefore ED' = GD' = DD', which means that the triangle $\triangle EDD'$ is isosceles with DD' = ED'. Since $DD' \parallel AB,$ we have $\angle DD'E = \angle ABC = 80^\circ$ and $\angle EDD' = \angle DED' = 90^\circ - \frac{\angle DD'E}{2} = 50^\circ.$ For the same reason, $\angle ADD' = 180^\circ - \angle D'DC = 180^\circ - \angle BAC = 100^\circ.$ Now the angle $\alpha = \angle AED$ can be found from the triangle $\triangle AED$ with the angles $\angle DAE \equiv \angle CAE = 10^\circ$ and $\angle ADE = \angle ADD' + \angle D'DE = 150^\circ$:

$\alpha = \angle AED = 180^\circ - (\angle DAE + \angle ADE) = 20^\circ$


group theory
abstract algebra
topology
advanced fields
advanced fields unsolved

Prove that if $ G$ is a topological group and $ H \leq G$ its closed subgroup, then $ G/H$ with quotient topology is $ T_{3}$.

$ G/H$ needn't be a group, because we do not assume that $ H$ is normal.

Oh... :blush:

But it is still not very complicated. A topological space $ (X,\tau)$ is $ T_3$ iff for every $ x\in X$ and $ V\in\tau$ with $ x\in V$ there is an $ U\in\tau$ with $ x\in U\subseteq\overline{U}\subseteq V$.

Let $ p: G\minus{}>G/H$ the canonical map. By the definition of the quotient topology p is continuous, open and closed. Now let $ V\subseteq G/H$ an open set containing the point $ xH$. Because G is $ T_3$ and p is continuous there is an open set $ U$ with $ x\in U\subseteq\overline{U}\subseteq p^{\minus{}1}(V)$.
$ \Rightarrow p(x)\in p(U)\subseteq p(\overline{U})\subseteq p(p^{\minus{}1}(V))\subseteq V$
$ \Rightarrow xH\in p(U)\subseteq \overline{p(U)}\subseteq p(\overline{U})\subseteq V$ because p is closed.

The set $ p(U)$ is open, because p is open. Therefore $ G/H$ is $ T_3$.

$ H$ closed means that $ G/H$ is hausdorff. no matter if the subgroup $ H$ is closed or not, the space $ G/H$ is regular (in particular, $ G / \{1\} \cong G$ is regular):

first observe that $ p : G \to G/H$ is open and that $ G$ acts continuously on $ G/H$. given $ x \in G/H$ and $ F \subseteq G/H$ closed such that $ x \notin F$, we have to separate $ x$ and $ F$. we may assume $ x\equal{}1H$. then $ F^c$ is an open neighborhood of $ 1H$ and since $ 1 \equal{} 1(1H)$ we find open neighborhoods $ 1 \in U \subseteq G$ and $ 1H \in V \subseteq G/H$ such that $ UV \subseteq F^c$. now let's see $ \overline{V} \subseteq F^c$: take $ v \in \overline{V}$. since $ U^{\minus{}1} v$ is an open neighborhood of $ y$, it intersects $ V$. thus we find $ u \in U$ such that $ v \in uV \subseteq F^c$. thus we have shown $ \overline{V} \cap F \equal{} \emptyset$ and $ x,F$ are separated by $ V,\overline{V}^c$.

what about a generalization? a topological group $ G$ acts on a topological space $ X$. do we have $ Gx \cong G/G_x$ (then in particular, the orbit $ Gx$ is regular)? in other words, is $ G \to Gx, g \mapsto gx$ open?


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $ H$ be a proper subgroup of group $ G$. Is it true that for any $ x$ in $ H$ there exists $ a,b$ in $ G$ \ $ H$ s.t. $ x\equal{}ab$.

no it is not
Take $ G \equal{} \mathbb Z$, $ H \equal{} 2 \mathbb Z$, $ x\equal{}2$.

We should find two odd numbers $ a,b$ such that their product is 2, in particular even. That is false because would mean that $ 2 | ab$ but $ 2 \not | a$ and $ 2 \not | b$ which is not possible since 2 is prime in $ \mathbb  Z$

bye

Think about the meaning of "group" again  :wink:

[color=red]If $ G$ is finite and $ \vert H \vert < \frac {\vert G \vert}{2}$ the result must of necessity be true, though... A rather well-known result, right?[/color] :D

That thing you probably point to is a lot harder than this completely trivial problem here  :D

Might be harder, but there is no denying it is also very interesting, ay?

Ok..I see it's false,sorry for posting it without thinking more of a counterexample, but I found it in the solution of this problem(which maybe it's not false):

[i]Let $ G$ be a group, $ H$ a proper subgroup and $ f: G\minus{}>G f(xy)\equal{}f(x)f(y)$ for all $ x,y$ in $ G$ \ $ (H$ \ $ {e})$. Prove that $ f$ is morphism.[/i]

$ x \equal{} a(a^{\minus{}1}x)$, for any $ a$ not in $ H$, $ x \in H$.

:blush:  :blush:  I got confused by the first example with $ Z$ and $ 2Z$ and so I concluded it's false...

But at least coquitao's problem is interesting in this topic :P


vector
linear algebra
linear algebra unsolved

Let E be a Vector Space and let $( | )$ be a positive, self-adjoint, sesquilinear form (inner product) on E.

And $N = \{x \in E | (x|x) = 0 \}$

Then on $E^{\ast} = E/N$, the map defined by $(x^{\ast}|y^{\ast}) = (x|y)$, where $x^{\ast} = x + N$ and $y^{\ast} = y + N$ is an inner product.

-----------

I'm not looking for the complete solution, i'm just confused about the definition of the inner product. So if someone could help me show one of the properties of an inner product that would be great. Thanks.

"semipositive" might be a better word than "positive" in this case.

[quote="julien_santini"]In the first definition you put "inner product" into brackets, but an inner product needs the condition $(x,x)=0 \Leftrightarrow x=0$, which is the reason why you factor $E$ by $N$ afterwards.[/quote]

Ok.

But how do I show, for example, that $(x^{\ast}|y^{\ast}) \geq 0$ ?

You don't; that's not part of any inner product definition.

My mistake, I meant: $(x^{\ast}|x^{\ast}) \geq 0$


limit
calculus
derivative
calculus computations

I was just trying to solve this problem: "Prove that if $lim(x_n)=a$ and $lim(t_1+t_2+t_3+...t_n) = \infty$  $t_i>0.$
So  $lim \frac{t_1x_1 + t_2x_2+...+t_nx_n} {t_1+t_2+t_3+...t_n}=a$
I got this solution(really "new" for me) and i want to know if this is correct:
For a given $\epsilon>0$. Take $n>n_0$  for wich $(x_n-a)<\epsilon$.

Look for $|\frac{\sum_{i=n}^{\infty} t_ix_i}{\sum_{i=n}^{\infty} t_i} -a| = |\frac{\sum_{i=n}^{\infty} t_ix_i -a\sum_{i=n}^{\infty} t_i}{\sum_{i=n}^{\infty} t_i}|$
$=|\frac{\sum_{i=n}^{\infty} t_i(x_i -a)}{\sum_{i=n}^{\infty} t_i}| < \frac{\sum_{i=n}^{\infty} t_i(\epsilon)}{\sum_{i=n}^{\infty} t_i}| <\epsilon$
So, this complete the proof??? I can "discard" finite(n_0) terms of the sum, no??
thanks!

But from $\lim_{n \infty} t_1 +t_2+ \cdots + t_n = \infty$ It seems that $\sum_{i=n}^{\infty} t_i$ divergen, so you can't do such arithmatic in your proof.

The  Césaro's theorem state more general

Let \[ a_0, a_2, \cdots , a_n, \cdots; \qquad b_0, b_2, \cdots , b_n, \cdots  \]

Be the infinite sequence satisfy the following conditions:
\[ b_n > 0 \qquad n = 0, 1, 2 , \cdots  \]
\begin{eqnarray*}\sum_{i=0}^{\infty} b_i \, t^i &\text{is convergen for}& |t|< 1 \\ &\text{is divergen for}& t=1 \end{eqnarray*}

\[ \lim_{n \infty} \frac{a_n}{b_n} = L  \] then
\[ \sum_{i=0}^{\infty} a_i t^i \qquad \text{is convergen for} \, |t|<1  \]
and
\[ \lim_{t \rightarrow 1-0} \frac{\displaystyle \sum_{i=0}^{\infty} a_i t^i}{\displaystyle \sum_{i=0}^{\infty} b_i t^i} = L.  \]

so that arithmatic is false??? but in $lim \frac{2n^2+n+1}{n^2} = 2$ we use the same arithmetic "cutting" n^2... but $lim n^2 = \infty$
tkz!

You used $\epsilon - \delta$ argument in your proof. That is not in limit form, You did the  arithmatic without limit sign.
In $\lim_{n \rightarrow \infty} \frac{2n^2 + n +1}{n^2}$ we can do the arithmatic because there's $\lim$ sign.

Ok! i will try to find antoher proof...
tkz

Igor, an easy and cool way of doing this problem is to use [url=http://en.wikipedia.org/wiki/L%27Hopital%27s_Rule]L'Hopital's Rule[/url]

substituting \infty into the equation right way would yield the indeterminate form of (infinity/infinity). So you would take the derivative of the num. and den. to get
lim n--->infinity (4n + 1)/(2n)

substituting \infty would get the same indeterminate form. So you take the derivative again of the num. and den. to get lim n--->infinity (4/2)

which simply equals 2. Nifty huh, I love L'Hopital's Rule but my classmates seem to dread it, so my teacher is usually forced to use longer methods to obtain some results.

Hope this helps


vector
linear algebra
linear algebra theorems

1) Prove that any algebra automorphism of the set of linear maps on $V$ is interiour. Here $V$ has finite dimension.
  2) Prove the same result without the assumption that $V$ is finite dimensional.

Ok, since I'm very proud I have finally found a solution for b), I will post it with the risk of spoiling the problem. I would have never found this solution without Leva's great solution for maps conserving rank.
    Let $f$ be an algebra automorphism of $L(E)$.
   Let us say that $p\leq q$ for two projectors if $Ker q$ is a subset of $Ker p$ and $Im q$ includes $Im p$. This is the same as $pq=qp=p$ and also it is obviously an order relation on the set of projectors.

    [b]Lemma 1[/b]
    The set of minimal elements with respect to this relation is the set of projectors of rank 1.
    [b]Proof:[/b]
    Suppose $q$ has rank 1 and $p\leq q$.  Thus since $Im p$ is a non-zero space included in one of dimension 1 ($Im q$), we must have $p=q$ and $q$ is minimal. Now, assume that $rank(q)>1$ and let's prove we can find $p$ different from $q$ and smaller than $q$. Take $x$ different from $0$ in $Im q$ and write $Im (q)=<x>+F$ and define the projector $p$ by $Im(p)=F$ and $Ker(p)=<x>+Ker(q)$. It is clear that it is different from $q$ and smaller.

    [b]Lemma 2[/b]
      The image of a minimal projector by $f$ is a minimal projector.
    [b]Proof:[/b]
     It is clear by the relations $f(pq)=f(p)f(q)$ that $f$ preserves this relation of order and the proof is now obvious.

     [b]Lemma 3[/b]
      If $rank(g)=1$ then $rank(f(g))=1$
     [b]Proof:[/b]
       Take $p$ the projector on $Im(g)$. It is a minimal projector (it has rank 1) and so $f(p)$ is a minimal projector. But of course, $f(g)=f(p)f(g)$ and so $rank(f(g))\leq 1$. Since $f$ is injective, $rank(f(g))=1$.
       Now, do what Leva did in rank presevers problem. Define $(a*+x)(t)=a*(t)x$ for a linear form $a*$ and a vector $x$.  The following observation is pretty clear:

      [b]Lemma 4[/b]
       We can always assume that there are a linear form $a*$ and a vector $a$ such that $p=a*+a$ is fixed by $f$ and $a*(a)=1$.
      [b]Proof:[/b]
       Take $a*$ a linear form and $a$ such that $a*(a)=1$. Then $a*+a$ is a projector of rank 1, thus it is sent by $f$ in a projector of rank 1.  But observe that if we have two projectors of rank 1, $p,q$ then there is invertible $u$ with $p=u^{-1}qu$. Indeed, write $E=<x>+Ker(p)=<y>+Ker(q)$  and take $u$ with $u(y)=x$ and $U(Ker(q))= Ker(p)$ (to see that $u$ is invertible, you can construct its inverse by changing the role of $p,q$). Thus we have seen that we can find $u$ invertible with $f(a*+a)=u^{-1} (a*+a)u$. Considering $u^{-1}fu$, which is an algebra automorphism, we may thus assume that we have $a*,a$ with $a*(a)=1$ and $f(a*+a)=a*+a$.
      Assume that we can find a linear form $a*$ such that $a*(a)=1$ and denote $p=a*+a$. Take any $x$ different from 0. Then $a*+x$ is of rank 1, thus $f(a*+x)$ is of rank 1. I claim that $Ker(a*)$ is included in $Ker (f(a*+x))$. This follows from the fact that $f(a*+x)(Id-(a*+a))=0$. Thus, $Ker(a*)$ and $Ker(f(a*+x))$ are hyperplanes   which contain one another, thus are equal and we have the existence of a vector $g(x)$ with $f(a*+x)=a*+g(x)$.  Obviously, it is unique and even more, it depends linearly of $x$. Reasoning with $f^{-1}$ we also see that $v$ is invertible. Easy computation shows that $a*+( v^{-1}f(t) v(x))= a*+ f(x)$ for all $x$ and all maps $t$. Thus $v^{-1} f(t) v=t$ and it's finally done.


Could someone please post the code that they would use to make a good looking image of the first couple of rows of Pascal's Triangle using LaTeX? Thanks.

The [b]\tabular[/b] environment can be used to center everything nicely, but it's quite tedious to do:\[ \begin{tabular}{ccccccccc} & & & & 1 & & & & \\
& & & 1 & & 1 & & & \\
& & 1 & & 2 & & 1 & & \\
& 1 & & 3 & & 3 & & 1 & \\
1 & & 4 & & 6 & & 4 & & 1\end{tabular}
\][code]$$\begin{tabular}{ccccccccc}
&&&&1&&&&\\
&&&1&&1&&&\\
&&1&&2&&1&&\\
&1&&3&&3&&1&\\
1&&4&&6&&4&&1\end{tabular}$$[/code]
You could make a program to generate the first $ n$ rows... :)

Thanks. I think a program would be overkill - I only need about five rows. :)

[quote="Twiz"]I only need about five rows. :)[/quote]
Then you have the code already. :D


$4kl+2k+2l+1=(2k+1)(2l+1)$ which is a square whenever $k=l$

Anyways, consider the following.

Set $b^2-4ac=(2y+1)^2-4(2x+1)(2z+1)$

We get $4(y^2+y)-3+8(2xz+x+z)$, so that $b^2-4ac\equiv 4(y^2+y)+5\bmod 8$

But $y^2+y\equiv 0\bmod 2$, so that $4(y^2+y)\equiv 0\bmod 8$.  Hence $b^2-4ac\equiv 5\bmod 8$.  But no perfect squares are $5\bmod 8$

This establishes the result.

[quote="spx2"]having $ax^2+bx+c=0$ and $a,b,c$ odd proove that $x$ is not in $Q$..[/quote]
The irreductible fraction  $x=\frac uv$   can't be solution because :$u^2+uv+v^2=1(mod \ 2) $
 :cool:

from where do you know that no square $k^2(mod 8)=5$

From evaluating all possibilities $\mod 8$ I think...

[quote="spx2"]diogene what you say makes no sense really
_________________
sunt prost
[/quote]

Tell me man, why do you say that about you self ?
  :cool:

By:
$0^2 \equiv 4^2 \equiv 0 \mod 8$,
$1^2 \equiv 3^2 \equiv 5^2 \equiv 7^2 \equiv 1 \mod 8$,
$2^2 \equiv 6^2 \equiv 4 \mod 8$
(I know there is an even simpler way but...)

[quote="spx2"]having $ax^2+bx+c=0$ and $a,b,c$ odd proove that $x$ is not in $Q$.

somewhere in the proof i have to do this $(\forall )(k,l) \in Z^2$ the expression $4kl + 2k+2l+1$ is never a perfect square.[/quote]

this is another solution
there are two kinds of the solutions:  x =p/q   p and q are odd,      or x=p/q  where p Or q is odd, the other is even.

case1:
ap²/q²+bp/q+c=0 ==>  ap²+bpq+cq²=odd+odd+odd=0 (even)   is not true.

case2: p Or q is even
ap²+bpq+cq² =0 gives  even+even+odd =0 Or odd+even +even=0  false too.

good jov various!
i did the same think
it was one of the problems in an old moroccan tst


I did a search here for \multline and the dialog seemed to be indeterminate.  Even this simple example:

[code]\begin{multline}
a+b+c+d+e+f\\
+i+j+k+l+m+n
\end{multline}[/code]

does not seem to work here or in TeXnicCenter.  Guess I can't get the code below formatted as a "multi-line"?  I think it would look better if the second line was shifted some to the left to give a better indication of a product or no?
\begin{align*} \ds\frac {ze^{\pi i/z}}{(1 - z)^2(2 + \pi iz)} = \bigg[ \frac {z}{2} + \left(1 - \frac {i \pi }{4}\right) z^2 + \frac {1}{8} \left(12 - 4 i \pi - \pi ^2\right) z^3 + \frac {1}{16} i \left( - 32 i - 12 \pi + 4 i \pi ^2 + \pi ^3\right) z^4 + \cdots\bigg] \\
\bigg[1 + \frac {i \pi }{z} + \frac {1}{2} \left(\frac {i \pi }{z}\right)^2 + \frac {1}{6} \left(\frac {i \pi }{z}\right)^3 + \frac {1}{24} \left(\frac {i \pi }{z}\right)^4 + \cdots\bigg] \end{align*}
Any ideas?
Thanks,
Shaw

. . . great . . . now it doesn't fit in the screen

You need the split environment rather than multline. In a document you would put
[code]\begin{equation*}
\begin{split}
\frac {ze^{\pi i/z}}{(1 - z)^2(2 + \pi iz)} &= \left[ \frac {z}{2} + \left(1 - \frac {i \pi }{4}\right) z^2 + \frac {1}{8} \left(12 - 4 i \pi - \pi ^2\right) z^3 + \frac {1}{16} i \left( - 32 i - 12 \pi + 4 i \pi ^2 + \pi ^3\right) z^4 + \cdots\right] \\
& \quad \times \left[1 + \frac {i \pi }{z} + \frac {1}{2} \left(\frac {i \pi }{z}\right)^2 + \frac {1}{6} \left(\frac {i \pi }{z}\right)^3 + \frac {1}{24} \left(\frac {i \pi }{z}\right)^4 + \cdots\right]
\end{split}
\end{equation*}[/code]
but in the forum you need to use align* rather than equation*
\begin{align*}\begin{split} \frac {ze^{\pi i/z}}{(1 - z)^2(2 + \pi iz)} & = \left[ \frac {z}{2} + \left(1 - \frac {i \pi }{4}\right) z^2 + \frac {1}{8} \left(12 - 4 i \pi - \pi ^2\right) z^3 + \frac {1}{16} i \left( - 32 i - 12 \pi + 4 i \pi ^2 + \pi ^3\right) z^4 + \cdots\right] \\
& \quad \times \left[1 + \frac {i \pi }{z} + \frac {1}{2} \left(\frac {i \pi }{z}\right)^2 + \frac {1}{6} \left(\frac {i \pi }{z}\right)^3 + \frac {1}{24} \left(\frac {i \pi }{z}\right)^4 + \cdots\right] \end{split}\end{align*}
Look at [url=http://tug.ctan.org/tex-archive/info/math/voss/mathmode/Mathmode.pdf]Math Mode[/url] sections 25 & 26 for the full details of the various amsmath environments.

I obviously don't understand Big Business:  Seems if I enter the code without the dollar signs in TeXnicCenter:

[code]\begin{multline*}
\ds\frac{ze^{\pi i/z}}{(1-z)^2(2+\pi iz)}=\bigg[
\frac{z}{2}+\left(1-\frac{i \pi }{4}\right) z^2+\frac{1}{8} \left(12-4 i \pi -\pi ^2\right) z^3+\frac{1}{16} i \left(-32 i-12 \pi +4 i \pi ^2+\pi ^3\right) z^4+\cdots\bigg] \\
\bigg[1+\frac{i \pi }{z}+\frac{1}{2} \left(\frac{i \pi }{z}\right)^2+\frac{1}{6} \left(\frac{i \pi }{z}\right)^3+\frac{1}{24} \left(\frac{i \pi }{z}\right)^4+\cdots\bigg]
\end{multline*}[/code]

that splits the line nicely.
Thanks for the code above Steve.  However I may still need to modify it slightly as it does not fit on the screen without using arrow keys.

You need the \usepackage{amsmath} command to be able to use the multline environment in $ \text{\LaTeX}$. I don't think it is enabled on the forum. So, stick with the align or aligned when posting; it gives you more control anyway:

$ \begin{aligned} &\ds\frac {ze^{\pi i/z}}{(1 - z)^2(2 + \pi iz)}\\ &= \bigg[ \frac {z}{2} + \left(1 - \frac {i \pi }{4}\right) z^2 + \frac {1}{8} \left(12 - 4 i \pi - \pi ^2\right) z^3 + \frac {1}{16} i \left( - 32 i - 12 \pi + 4 i \pi ^2 + \pi ^3\right) z^4 + \cdots\bigg] \\ &\hspace{100pt}\bigg[1 + \frac {i \pi }{z} + \frac {1}{2} \left(\frac {i \pi }{z}\right)^2 + \frac {1}{6} \left(\frac {i \pi }{z}\right)^3 + \frac {1}{24} \left(\frac {i \pi }{z}\right)^4 + \cdots\bigg] \end{aligned}$


[img]http://www2.hkedcity.net/sch_files/a/lpl/lpl-lsk/public_html/cake2.gif[/img]
As shown in the figure, ABCD is a cake in the shape of a square with AB = 10 inches. The cake will be cut into pieces along straight lines  AE, BF, CG and DH. Besides, AH = BE = CF = DG.
If it is required that the area of square PQRS should be 50 square inches, how long should  AH be chosen?

Let $a = AB = 10\ \text{in}$ be the side of the square $ABCD$, $X \equiv AC \cap BD$ the intersection of its diagonals and $P', Q', R', S'$ the midpoints of the sides $AB, BC, CD, DA$, forming a square $P'Q'R'S$ the side $P'Q' = \frac{a\sqrt 2}{2}$ and area $S_{P'Q'R'S'} = \frac{S_ABCD}{2}$ equal to half the area of the original square $ABCD$. The desired square $PQRS$ is congruent to the square $P'Q'R'S'$. Let $(X, r)$ be the incircle of the square $P'Q'R'S'$ with radius $r = \frac{P'Q'}{2} = \frac{a\sqrt 2}{4}$. Rotate the square $P'Q'R'S'$ clockwise until its extended side $S'P'$, identical with the tangent to its fixed incircle $(X, r)$ at the midpoint $T'$ of the side $S'P'$ passes through the vertex $D$ of the square $ABCD$. From the right angle triangle $\triangle XTD$,

$\sin{\widehat{XTD}} = \frac{XT}{XD} = \frac{a\sqrt 2}{4} \cdot \frac{2}{a \sqrt 2} = \frac 1 2,\ \ \angle XDT = 30^o$

$\angle ADH \equiv \angle ADH = 45^o - \angle XDT = 30^o = 15^o$

From the right angle triangle $\triangle ADH$,

$AH = AD \tan{\widehat{ADH}} = a \tan{15^o}$

$\tan{30^o} = \frac{2 \tan{15^o}}{1 - \tan^2{15^o}} = \frac{\sqrt 3}{3}$

$\tan^2{15^o} + 2\sqrt 3 \tan{15^o} - 1 = 0$

$\tan{15^o} = \frac{-2\sqrt 3 \pm \sqrt{12 + 4}}{2} = \pm 2 - \sqrt 3$

Since $0 < 15^o < 90^o,\ 0 < \tan{15^o} < \infty$, hence $\tan{15^o} = 2 - \sqrt 3$ and $AH = a(2 - \sqrt 3) \doteq 2.68\ \text{in}$.


A point S is chosen inside the square MNPQ. What is the probability that the angle MSN is acute?

In order for angle MSN to be acute, S must fall out of the semicircle constructed inside the square with MN as the diameter. If the side of the square is $a$, the area of that semicircle is ${{({a \over 2})^2 \pi}\over 2}={a^2\pi \over 8}$. Therefore, the desired probability is ${{a^2-{a^2\pi \over 8}} \over a^2}=1-{\pi \over 8}$.

Why is the angle in semicircle obtuse?Although this is quite obvious.

A semi-circle can always be circumscribed around a right triangle - this means that given any two adjacent vertices of a square, any point on the inscribed semi-circle (with those two points forming the diameter) will be a right angle. Therefore, anything inside is obviously obtuse, and anything outside is acute.


Hi, I'm doing a physics project and my teacher told me to research about something called the magnus force. I looked it up, but I can't understand much of what on earth they are talking about. Can someone please explain it to me in a not-so-complicated way? Please? :?:

I think it is the force which acts on say a ball as it moves through a flluid. Try googgling!!

The Magnus Force is a force which acts on a rotating ball moving through a fluid. It causes the ball to curve - baseball / soccer / golf players use this a lot. The basic principle is that the spinning ball distorts the air flow around it - it makes the air flow around the ball slower on one side and faster on the other side, thus being pushing to the side where the air goes faster. (People like to explain this part by Bernoulli's Principle.) If you search Google Images and type in magnus force, you'll find a couple of nice pictures explaining this a little better.

The key to the magnus force is that as the object spins, the flow on one side has a lower velocity than the flow on the other side due to viscous forces between the fluid and the object. In flows, raising the velocity lowers the pressure, so the side with high velocity fluid has a lower pressure than the side with the low velocity fluid. This pressure difference is what causes "The Magnus Force."

Hope this helps.

how will it affect a ROLLING ball, though?

I don't think it does much to a rolling ball. It might push the ball into the table (or whatever it is rolling on) a little but probably not too much. It won't affect the ball's velocity - air resistance could, but that's not Magnus Force. It could also increase the friction a little bit because of the increased normal force but as I said, it would be probably so little that it's not really worth thinking about.


I found a quite nice solution for this problem:

Consider the sequence {un} defined by

u0 = 1, un + 1 = aun + b.

Prove that for any a, b, {un} contains infinitely many composites.

[GERMANY 1995]

However, I heard that another - more natural - solution exists.
So, can you find it?

I assumed a and b to be positive integers (I'm sure it works for integers a and b, but my solution only works if u_n is positive for large enough n's). Maybe it's not complete; I don't know what conditions to take for a and b since you didn't mention any, but I'm posting this because it's pretty short and nice:

It's easy to see that u_n = a^n+b*(a^n-1)/(a-1). Let's assume u_n is a prime for all n>=n0. We then look at the term u_(n*u_n) = a^(n*u_n) + b*[a^(n*u_n)-1]/(a-1). From Fermat's thm we have k^u_n = k (mod u_n), so a^(n*u_n) = a^n (mod u_n), so u_(n*u_n) = a^n+b*(a^n-1)/(a-1) (mod u_n) = u_n (mod u_n) = 0 (mod u_n), so u_(n*u_n) isn't a prime, but n*u_n > n >= n0, so we have a contradiction.

Like I said, this works for a and b s.t. u_n becomes large and, more importantly, positive, for large values of n.

I showed that if n = k\emptyset(a + b) + 1 for any k, then un is composite.

Yup! I see what you mean:

If we assume the contrary of what we want to prove, then (a,b)=1, so (a,a+b)=1, which means that a^ \emptyset (a+b) = 1, and if we plug
n=k* \emptyset (a+b)+1 into u_n = a^n + b*(a^n-1)/(a-1) we get exactly what you said: a+b | u_n, so u_n isn't prime.

However, I think my soln can still be saved. I think that even if a is negative (or b), we still get positive u_n 's for infinitely many large enough n's.

Well, in your continuation of the outline of my proof, there is a small error.

It is not necessarily true that (a + b)|un ... or am I wrong?

Now I'm in doubt myself :?

I think you're right. I'm not too sure, but just in case we could say this: p|u_n, where p is a prime which divides a+b.

I reached this:

(a + b)/d divides un, where d = gcd(a + b, a - 1).

Is that correct?

As stated the problem is false if a=-1 and b-1 is a prime. Then all the terms are 1,b-1,1,b-1,... so none of them is prime. So in order for the problem to be true we need a is not -1. (Or else if a can be -1 as well, taking b-1 composite solves the problem and b-1 prime contradicts it).

Of course, we need gcd(a,b)=1 or otherwise the problem is trivial.

If a=1 then we have u_{n+1} = u_n + b and by choosing a prime number p which does not divide b, we can easily find an infinite number of terms divisible by p.

If there is n such that |u_n| is prime and greater than a-1, let us fix n. Otherwise since u_n is unbounded, there can be only a finite number of prime terms in [-a+1,a-1]  \cap Z so u_n contains an infinite number of composites.

Let us work modulo u_n :

u_{n+1} = b
u_{n+2} = b(a+1)
..................
u_{n+k} = b(a^(k-1)+...+a+1) , for all k in N so :

So u_{n + k*u_n - 1} = b * (a^{k*u_n - 1} - 1)/(a-1) = 0 (mod u_n), because gcd(a-1,u_n)=1!  Using again the fact that u_n is unbounded we are done.

Now I've read all the solutions above and I noticed that grobber had pretty much the same idea. Sorry for posting the same solution all over!

By the way grobber, can you please post the solution to the problem about sets A and B s.t. A+A=B ,  A \cup B = R , A\cap B= \emptyset and A,B are dense in R?

Sure, but i'll have to look it up in one of my notebooks first.


(1) Prove that there exist five nonnegative real numbers $ a, b, c, d$ and $ e$ with their sum equal to 1 such that for any arrangement of these numbers around a circle, there are always two neighboring numbers with their product not less than $ \frac{1}{9}.$

(2) Prove that for any five nonnegative real numbers with their sum equal to 1 , it is always possible to arrange them around a circle such that there are two neighboring numbers with their product not greater than $ \frac{1}{9}.$

[quote="hmm"](1) Take a=b=c=1/3 and d=e=0. Clearly, two of a,b,c have to be adjacent in any arrangement, proving the claim.

(2) Two of the numbers must sum to <=2/3 (or else four of the numbers sum to >4/3, contradiction). By AM-GM, these two have a product of at most 1/9, so just take an arrangement where these two #s are adjacent.

(Was this really a #4 on CGMO? It seemed really easy...only took 5 min...)[/quote]

I checked the original Chinese version. Part (2) should read "... such that no two neighbouring numbers have a product greater than $\frac{1}{9}$.  :D

That doesn't make it much more difficult.

Assume $a\le b\le c\le d\le e$. Easy to prove that the ordering $(a,e,b,c,d)$ will do the job:

 As $ad\le cd,ae\le be,bc\le be$, we only have to show $be\le\frac 19$ and $cd\le\frac 19$.

Suppose $cd>\frac 19$. Then by AM-GM, $c+d>\frac 23$, thus $\frac 13<d\le e$, thus $c+d+e>1$ contradiction.

Suppose $be>\frac 19$, thus $3b>\frac 1{3e}$. From $3b+e\le b+c+d+e\le 1$, we get $\frac 1{3e}+e<1$ or $(e-\frac 12)^2<-\frac 1{12}$, again a contradiction.


induction
number theory unsolved
number theory

Let a and b be integers . Show that a and b have the same parity if and only if there exist integers c and d such that $a^2 +b^2 +c^2 +1=d^2$.

Here the link:
http://www.mathlinks.ro/Forum/viewtopic.php?t=6236&highlight=a%5E2%2Bb%5E2%2Bc%5E2%2B1%3Dd%5E2

the solution to this utilizes the fact that perfect squares can only be $0$ or $1 \bmod 4$ right?

Yes - I think so.

Firstly, we rearrange the equation to $a^2 + b^2 + 1 = (d+c)(d-c)$.

$a^2 + b^2 + 1 \equiv \{1,3\} \bmod{4}$ iff $a$ and $b$ are of the same parity
$(d+c)(d-c) \equiv \{0,1,3\} \bmod{4}$ (cannot be 2 because $(d+c)$ and $(d-c)$ have the same parity, so if both even, the product is congruent to $0$.

Therefore if $a$ and $b$ have the same parity there must exist an equation of the above form.

Now, if $a$ and $b$ have different parity, $a^2 + b^2 + 1 \equiv 2 \bmod{4} \notin \{0,1,3\}$, completing the proof.

Where have you shown that there exists $c$ and $d$?

Sorry, I don't know why I skipped the issue over originally.

There exists $(c,d)$ where $(c+d)(c-d) = 1$ (namely $(1,0)$).
Now, by induction, if we assume $(c+d)(c-d) = 2k+1\ (\forall k \in \mathbb{N})$, and $(c-d) = 1$, then $(c+d+2)(c-d) = 2(k+1)+1$, giving the new values $(c',d') = (c+1, d+1)$.

This covers [b]all[/b] values $1 \bmod{2} \equiv 1\bmod{4}$ or $3\bmod{4}$.