geometry
3D geometry
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binomial theorem

Can you beat 3min:16sec?

Rules: Do the problems(without a calculator), see how much time you took, and correct the problems. Add 5 seconds to your time for every wrong answer.

Here are the problems:

487+64
671-89
11478+1999
47*7
78*9
187*64
1111*2222
489/76
611/12
782/11

For the division problems, give the quotient and remainder, it's easier that way instead of putting down the decimal. :)

i did it in 4:16.

i didn't read the part about "just give the remainder."  i was going along with the decimal.

oh, well, it was fun.

[quote="1=2"]
Here are the problems:

487+64
671-89
11478+1999
47*7
78*9
187*64
1111*2222
489/76
611/12
782/11
[/quote]
these aren't problems
they dont require thinking
even though you could be cool and do stuff in easier ways

2
2
2
2
2
2
2
2
2
2

for a score of 53 seconds :D
i think there should be more penalty for wrong answers :rotfl:

3 minutes, 18.26 seconds, all answers correct.

@tjhance: I don't think you can claim that your time is accurate to the hundredth of a second, so it makes more sense to just round to the nearest second (unless you are some sort of time-keeping god :roll: )

I could see him using a rubic's cube timer on that...

w00t!  40 seconds, 1 wrong (sort of.  I skipped 489/76 to save time) for a time of 45 sec.

Oh, and for 1111x2222:
[code]
   1111
  1111
 1111
1111
-------
1234321
[/code]

Now double it.  Hmmm...

I especially liked the last one.

I asked my teacher to make a new one, with about the same difficulty, and I got 1 minute 6 seconds :)

I used the power-of-ten method of multiplication to get 1111x2222

1*2 = (2) *1
(1*20) + (10*2) = (2 + 2) * 10
(1*200) + (10*20) + (100*2) = (2+2+2) * 100
etc etc

add it up for 2468642

its basically one giant FOIL-type distribution that works well with small digits

I got 2:11, 1 wrong, for a total of 2:16.
Is that good for a 6th grader? (I'm in 6th grade)

Lol 52 seconds, 0 wrong. I was never any good with fast arithmetic...

The only particularly interesting problem there is $ 1111 \times 2222$, which suggests the following series of identities:

$ (1 \plus{} x)^2 \equal{} 1 \plus{} 2x \plus{} x^2$
$ (1 \plus{} x \plus{} x^2)^2 \equal{} 1 \plus{} 2x \plus{} 3x^2 \plus{} 2x^3 \plus{} x^4$
$ (1 \plus{} x \plus{} x^2 \plus{} x^3)^2 \equal{} 1 \plus{} 2x \plus{} 3x^2 \plus{} 4x^3 \plus{} 3x^4 \plus{} 2x^5 \plus{} x^6$
...
$ (1 \plus{} x \plus{} x^2 \plus{} ...)^2 \equal{} 1 \plus{} 2x \plus{} 3x^2 \plus{} 4x^3 \plus{} ...$

Note the symmetry of the coefficients.  The third identity with $ x \equal{} 10$ gives the desired value of $ 1111^2$.  The last identity can be generalized using the [url=http://mathworld.wolfram.com/BinomialTheorem.html]general binomial theorem[/url] to give

$ (1 \plus{} x \plus{} x^2 \plus{} ...)^r \equal{} (1 \minus{} x)^{\minus{}r} \equal{} \sum_{k\equal{}0}^{\infty} {\minus{}r \choose k} (\minus{}x)^k \equal{} \sum_{k\equal{}0}^{\infty} {k\plus{}r\minus{}1 \choose k} x^k$

This turns out to be the generating function of the [url=http://www.google.com/search?hl=en&q=balls-and-urns&btnG=Google+Search]balls-and-urns[/url] problem.  :)

Mine is 50, above average here

[hide]I answer 0 for all the questions :D
To 1=2, i dont think one can solve each question in 5 second, so skipping may be a good strategy.
[/hide]

[quote="Ihatepie"]how did u do 1111 x 2222 so fast?

I did it in 1:30 thios time after forgettin answers[/quote]
1111*1111*2
trick for 1111111...(n 1's) squared fast is 12345678...n...654321 like for 1111 would be 1234321
then multiply by 2
i did it in 2:30 or less no mistakes but i forgot to keep track of time... xD

50 seconds. I just gave wrong answers to all of them. instantly.

(this is just to point out that gaping hole in the process)

Anyways, my actual time is 1 min, 16 seconds. Hmmm.

Meh, 59 seconds. No wrong. They're not that hard, but I used to pwn at these kind of problems in elementary school... guess I'm just horribly rusty.

If i can use an abacus i can get them pretty fast like 55 secs are abaci allowed?

2:12
Is that bad for a 6th grader?(I'm a 6th grader.)

As for the penalty:

(my idea, don't know if 1=2 would agree)

For every wrong answer you get, double your time once at the end (after adding the original penalty). (meaning if you just skipped all ten, you would get a time of 14 hours, 13 minutes, 20 seconds.  :rotfl:  :rotfl:  :rotfl:  :rotfl: )

EDIT: I think.

Eh, I got 1:14 (no wrong).  I should have done the multiplication more efficiently.

[quote="mathfreak77"]43 seconds with mental math (abacus in your head)
52 seconds without mental math. :lol:[/quote]
What is abacus? Oh, I got 50 seconds (0+10*5  :rotfl: ).

It's a manual Chinese instrument which speeds up calculation. :D

38 seconds - 5 right, 4 skipped, 1 wrong :D

It should be +15 seconds for wrong and skipped instead of just 5.

How do you guys calculate so fast?

Mewtwo... Just what are you?! How did you guys become so good at calculating so fast? Is there some kind of book or website that helps?


floor function
number theory
least common multiple
LaTeX

A machine for making donuts produces plain donuts. It then puts chocolate coating on top of every 24th donut, fills every 30th donut with strawberry jam and covers every 45th donut with sprinkles. Since the machine started working, it has processed 25000 donuts.

i) How many chocolate coated donuts with jam and sprinkles has the machine produced?

ii) How many donuts with jam that are either chocolate coated without sprinkles or with sprinkles but without chocolate coating has the machine produced?

Can someone check my symbols to see if I'm doing them right?

$ \displaystyle C= \{ x|24c$ for some c$ \} $
$ \displaystyle J= \{ x|30c$ for some c$ \} $
$ \displaystyle S= \{ x|45c$ for some c$ \} $

i) We are looking for $ \displaystyle x \in (S \cap J \cap C)$, $ \displaystyle x<25000$, which can be found by $ \displaystyle \lfloor \frac{25000}{lcm(24,30,45)} \rfloor = \lfloor \frac{25000}{360} \rfloor = 69$

ii) We are looking for  $ \displaystyle x \in ((J \cap C) - (J \cap S \cap C)) \cup ((J \cap S) - (J \cap S \cap C))$, $ \displaystyle x<25000$, which is found by $ \displaystyle (\lfloor \frac{25000}{lcm(30,45)} \rfloor - \lfloor \frac{25000}{lcm(24,30,45)} \rfloor) + (\lfloor \frac{25000}{lcm(30,24)} \rfloor - \lfloor \frac{25000}{ lcm (24,30,45)} \rfloor) = (277 - 69) + (208 - 69) = 208 + 139 = 347$

a) Yeah you're correct.  It's int[25000/lcm(24, 30, 45)].
b) Yeah that's correct as well.

Ahhh, so int is the command. Thanks for that.

Well that's just the general abbreviation for it, or the way you put it works as well.  I have no idea how to do the thing u did with latex.

K81o7: If you click on the LaTeX graphics, a small window will pop up with the TeX commands that created the graphic.


If B is the point (5,5) on the coordinate plane, then find the number of possible lattice points A such that 0<AB<=4.

EDIT: A lattice point is a point that has integer coordinates.

[hide]

don't know how to prove it but is it 48?

(I'm assuming that it means less than or equal to 40.[/hide]

haven't we had this problem already? Just draw the circle with radius 5.  Translate the circle to the origin as it won't affect the total number.

You mean radius 4?

[hide="Solution"]Consider a circle with radius-4 centered at the origin.

To count the total lattice points, we will consider the lattice points in Quadrant I, and multiply it by four since there are four quadrants. (We will deal with the points on the axes later.)

When $ x\equal{}1$, there are $ 3$ lattice points.
When $ x\equal{}2$, there are $ 3$ lattice points.
When $ x\equal{}3$, there are $ 2$ lattice points.

In total, there are $ 3\plus{}3\plus{}2\equal{}8$ lattice points in the First Quadrant.

On the axes, there are $ 4(4)\equal{}16$ lattice points.

Summing them all up, we have $ 4(8)\plus{}4(4)\equal{}4(12)\equal{}\boxed{48}$ lattice points.[/hide]


Solve the following systems of equations
x-y+z-w=2
x^2-y^2+z^2-w^2=6
x^3-y^3+z^3-w^3=20
x^4-y^4+z^4-w^4=66

isnt this the standard sort of question with viete's theorem+recursion?

I've run a couple of internet searches to try to ascertain the link between Vieta's theorem, recursion, and simultaneous equations but drawn a blank.
Could you explain in more detail what is the standard approach to which you refer?!

If they were pluses for the operations, you could use a sort of "reverse-Newton's sums"...maybe you could adapt this to the problem...

I don't think we can use some theorem to solve this problem.
I knew a nice solution:
Let $x'=x-\frac32,y'=y-\frac32,z'=z-\frac32,t'=t-\frac32$

i) Correction: read $yw=q$ .
ii) More generally if we consider the system:
$x-y+z-w=a,x^{2}-y^{2}+z^{2}-w^{2}=b,x^{3}-y^{3}+z^{3}-w^{3}=c,x^{4}-y^{4}+w^{4}-z^{4}=d$ let $\delta=-3b^{2}-a^{4}+4ac$.
a) If $\delta\neq0$ (here $\delta=36$) then there exists 1 solution for $s,t,p,q$.
For example: $t=\frac{3ab^{2}-2a^{3}b-4bc+6ad-4a^{2}c+a^{5}}{2\delta}$.
b) If $\delta=0$ there exist 0 or an infinity of solutions.
For example let $a=b=c=1$.
If $d=1$ then there exist an infinity of solutions: $s=t+1,p=t,q=0$.
If $d\neq1$ then there exist 0 solution.

Thank you foe explaining for me. That is very nice way


Let's discuss the solutions.

Gold 1. I think my approach is wrong.
Gold 2. n^3 * lgn solution. (matrix power)
Gold 3. nlgn solution (greedy)

I solved 1 with a DP:
let f(n, h) be the minimum cost for n poles with the n-th pole having height h.
then f(n, h) = min over all h' { (H[n]-h)^2+f(n-1, h')+C | h'-h | }
this is O(NH^2), but with some trickery you can cut it down to O(NH).

Anyways, I thought this contest was definitely harder than the ones from last year.

i got the nh^2 solution already, but i wonder what are the trick to make it nh?

Drat, that was nasty. I got all of 7 test cases on relays and 1 on each other.

I came up after the contest with a fibonacci-number solution to relays, which still doesn't run in time and barely runs in memory, but is awesomely cool to me. Let $ K_{(i,j)}^n$ be the cost to get to intersection $ i$ from intersection $ j$ with $ n$ cows, and $ I$ be the number of intersections. For $ n$ equal to the $ p$th fibonacci number 1,2,3,5,8,... $ K_{(i,j)}^n$ is the minimum of $ t \equal{} 1,2,\cdots,I$: $ K_{(i,t)}^{n \minus{} 2} \plus{} K_{(t,j)}^{n \minus{} 1}$. Then we simply represent $ N$ as a sum of fibonacci numbers and if $ p$ is in this sum then we store that one in memory. It requires 1 int to remember $ K_{(i,j)}$ for a given $ n$, and we have to remember 3 n to calculate it, so it requires 6MB memory, plus another set of them that we're remembering for the answer, for 8MB memory. You'll never need to calculate more than 24 $ n$ so worst case the actual $ k$-calculation takes $ 2\cdot 10^10$ additions, at least as many comparisons, ..., bleh. Obviously it doesn't fit in time, but it's cool IMO :).

EDIT: Wow that above is useless. Why couldn't you just do powers of 2 instead of fibonacci numbers? That would make it way faster, and we wouldn't even have to do additions, just <<1s. Is this $ O(I^2 \log n)$?

On telewire, the best I could come up with was running through the poles, and if a pole was a local minimum trying to increase the set of contiguous poles with that height to minimize the difference. [Trying to make the increase-in-height cost \leq 2*C]. Requires about 9 runs to minimize the difference, plus the time to figure out how long the run of equal-height poles is. Is that O(nh)? [I am awful at O-time calculation... I think there are $ \log h$ runs each with an average of $ n\log h$ changes, so nh is the worst case, $ n\log h$ best case?] Neal Wu said there was a dp way to do this one, but I don't understand it.

[code]
                     -- case number --
           1  2  3  4  5  6  7  8  9 10 11 12 13
relays      *  *  *  *  *  *  t  t  t  t  t  t
tanning     *  x  t  t  t  t  t  x  x  t  t  x  t
telewire    *  *  *  x  x  x  x  x  x  x
[/code]
Modified Bellman-Ford on relays - O(NT), saw an O(I^3 log N) solution at 20 minutes left but then realized that it was too slow and didn't notice the easy reduction to O(T^3 log N).
Incorrectly coded max flow on tanning - bad
O(NH) dp after an incorrect optimization on telewire - less bad, but still bad

[code]relays      *  t  t  t  t  t  t  t  t  t  t  t
tanning     *  t  t  t  t  t  t  t  t  t  t  t  t
telewire    *  *  *  *  *  *  t  t  t  t[/code]

I think i did pretty much the same dp as the previous guy.


For what (composite) positive integers $ k$ is it true that $ n^{\phi(k)\plus{}1}\equiv n\pmod{k}$ for all integer $ n$?

EDIT: k, not n :blush:

$ n\equal{}2$ gives $ 2^{1\plus{}1}\equiv 2(mod\ k)$ so $ k\equal{}1$ or $ k\equal{}2$ but here $ k$ is not composite
then there is no such composite integre

Perhaps he meant $ n^{\varphi(k)\plus{}1}$. In this case, I'm not quite sure of the answer, since for $ k\equal{}6$,

$ 0^{3}\equiv 0$
$ 1^{3}\equiv 1$
$ 2^{3}\equiv 2$
$ 3^{3}\equiv 3$
$ 4^{3}\equiv 4$
$ 5^{3}\equiv 5$, but for $ k\equal{}8$

$ 2^{5}\equiv 0\not\equiv 2$.

\[ n^{\varphi (k)\plus{}1}\equiv n\pmod{k}\]
when $ k$ is squarefree, since the congruence holds modulo every prime divisor of $ k$. Conversely, if $ k$ is not squarefree but divisible by $ p^{2}$, the congruence $ p^{\varphi (k)\plus{}1}\equiv p\pmod{k}$ is clearly false.


Find the number of ways one can cover a 3 x 8 grid with 12 identical dominoes.

I didn't understand the solution. Any ideas?

2X1
so you know how dominoes have 2 squares on them, each square is 1X1

I would suggest trying some smaller cases

I feel as though we need to figure out the number of possibilities recursively, but I can't seem to figure out exactly how...

Yeah, you can do it by recursion, it just needs some careful counting.
[hide="Hint"]
Probably the trickiest bit is working out what to do with this case:
[code]
A
A
BB
[/code]

[hide]
Either you fill in the gap with a vertical domino and get back to a rectangle left, or you fill it in with horizontal dominoes and get back to a smaller case 2.
[/hide]
[/hide]

TripleM,

[hide]I am confused about what you are saying. Are AA and BB two different dominoes? If the board you were envisioning this:

[code]
A00
A00
BB0
000
000
000
000
[/code]

Then you can fill in the 4x4 grid in the upper right with either two horizontals or two verticals.

I can see that dominoes must be placed in 3 x 4 blocks, but I'm still not sure how to find the recursive formula...
[/hide][/code]

Sorry, I had the grid oriented the other way.
Heres all the cases:
[hide]

Case 1:
[code]
AA...
BB...
CC...
[/code]
(AA being one domino etc).

Case 2:
[code]
A...
A...
BB..
[/code]
We can either put the next domino to fill up that 2x3 rectangle
[code]
AC...
AC..
BB..
[/code]
otherwise we must have
[code]
ACC...
ADD...
BBEE..
[/code]
where we can now repeat the procedure (either fill in the rectangle, or place another 3 horizontal dominoes).

The last case is just the above case upside down, so we have to double that.
[/hide]

Ok...so given that pattern, how would you develop the observations into a recursion f(n) for the number of ways to cover a 3 x n board with 2 x 1 dominoes?

[hide]
$f(n) = \underbrace{f(n-1)}_{\text{ case 1 }} + (\underbrace{f(n-1)+f(n-2)+f(n-3)+\ldots}_{\text{ case 2 and 3 }})\times 2$
[/hide]

I hope.

[hide]Well, case 1 would be f(n-2). I do not understand your reasoning for coming up with the expression for case 2 though (and it does not match the one given in the solutions file).[/hide]

Hmm..
[hide]
I let f(n) be the number of ways of tiling an 3x(2n) board, maybe thats what confused you. Then case 1 is f(n-1), and:
[code]
||..
||..
--..
[/code]
f(n-1) ways.
[code]
|--|..
|--|..
----..
[/code]
f(n-2) ways.
[code]
|----|..
|----|..
------..
[/code]
f(n-3) ways.
Etc etc.
[/hide]


real analysis
ratio
calculus
calculus computations

Determine whether the series converges or diverges.

$ \sum_{n\equal{}1}^{\infty}\frac{n^3(\sqrt{2}\plus{}(\minus{}1)^n)^n}{3^n}$

$ \sqrt{2}\plus{}1<3,$ so $ \left(\frac{\sqrt{2}\plus{}1}{3}\right)^n$ goes to zero geometrically fast. The $ n^3$ is trivial against that, so the series converges.

For a more formal argument, look at the limsup version of the root test:

$ \limsup_{n\to\infty}\left[\frac{n^3(\sqrt{2}\plus{}(\minus{}1)^n)^n}
{3^n}\right]^{\frac1n}\equal{}\frac{\sqrt{2}\plus{}1}3<1,$ so the series converges absolutely.

You do want to use the limsup version of the root test rather than the simpler limit version, and you don't want to use the ratio test at all.


But all the primes $ p \equal{} 4k \plus{} 1$ can be written $ p \equal{}a^2\plus{} b^2$ for $ a,b$ intergers  :D

[quote="pelao_malo"]But all the primes $ p \equal{} 4k \plus{} 1$ can be written $ p \equal{} a^2 \plus{} b^2$ for $ a,b$ intergers  :D[/quote]How to prove it?

[quote="Taidoka-X"]For every a there exist b such that a^2+b^2 is a prime. a and b are integers[/quote]

How i could prove this?

Fermat Christmas theorem cannot help to solve this problem. This problem is a partial case of famous open problem about prime values of polinoms. As far as I know,  some people try to prove that this hypothesis is unprovable.


Hello,

Could anyone recommend a problem book on integration, or a book on integration techniques with lots of examples? I am specifically looking for problems that are solved with a combination of conventional techniques and problem-specific tricks that are hard to think of. I could also use problems on limits. I'm not looking for calculus texts.

Thanks.  :)

[url=http://www.amazon.com/Table-Integrals-Products-Sixth-Gradshteyn/dp/0122947576]This[/url] isn't exactly what you want but it's a well-known and comprehensive book.  Although it doesn't have problems, it has LOTS of identities that you can use to solve problems.

Try [u]Schaum's Outline of Advanced Calculus[/u] by Wrede and Spiegel, or (the chapters dedicated to calculus from) [u]Problem Solving Through Problems[/u] by Loren Larson. Previews of each can be found on http://www.books.google.com

Thanks for the suggestions, I think "Problem Solving through Problems" by Larson is what I'm looking for.

Btw, there is a free integral table/formulae book from books.google.com [url=http://books.google.com/books?id=pYMRAAAAYAAJ&printsec=frontcover&dq=integrals&as_brr=1]here[/url].


$ x,y,z > 0$.prove that:

$ (i)$ $ \frac{x^2}{y^2} \plus{} \frac{y^2}{z^2} \plus{} \frac{z^2}{x^2}$ $ \geq$ $ \frac{y}{x} \plus{} \frac{z}{y} \plus{} \frac{x}{z}$

$ (ii)$ $ \frac{x^2}{y^2} \plus{} \frac{y^2}{z^2} \plus{} \frac{z^2}{x^2}$ $ \geq$ $ \frac{x}{y} \plus{} \frac{y}{z} \plus{} \frac{z}{x}$

Please use chebyshev inequality to solve this problem!! :)

[quote="behdad.math.math"][quote="hedeng123"][quote="behdad.math.math"]$ x,y,z > 0$.prove that:

$ (i)$ $ \frac {x^2}{y^2} \plus{} \frac {y^2}{z^2} \plus{} \frac {z^2}{x^2}$ $ \geq$ $ \frac {y}{x} \plus{} \frac {z}{y} \plus{} \frac {x}{z}$

$ (ii)$ $ \frac {x^2}{y^2} \plus{} \frac {y^2}{z^2} \plus{} \frac {z^2}{x^2}$ $ \geq$ $ \frac {x}{y} \plus{} \frac {y}{z} \plus{} \frac {z}{x}$

Please use chebyshev inequality to solve this problem!! :)[/quote]

AM-GM Inequality can be solved.[/quote]

Please solve it bye chebyshev inequality.
would you please write your solution by AM-GM?[/quote]
By the AM-GM inequality:
$ (i)$ we have $ \frac {{x^2 }}{{y^2 }} \plus{} \frac {{z^2 }}{{x^2 }} \ge 2\frac {z}{y}$ , $ \frac {{x^2 }}{{y^2 }} \plus{} \frac {{y^2 }}{{z^2 }} \ge 2\frac {x}{z}$
and $ \frac {{z^2 }}{{x^2 }} \plus{} \frac {{y^2 }}{{z^2 }} \ge 2\frac {y}{x}$
Adding up these three inequalities, we have the desired inequality
$ (ii)$
$ \frac {{x^2 }}{{y^2 }} \plus{} 1 \ge 2\frac {x}{y}$
$ \frac {{y^2 }}{{z^2 }} \plus{} 1 \ge 2\frac {y}{z}$
and   $ \frac {{z^2 }}{{x^2 }} \plus{} 1 \ge 2\frac {z}{x}$
But we also have $ \frac {x}{y} \plus{} \frac {y}{z} \plus{} \frac {z}{x} \ge 3$
So the ine is proved

Thanks for your solution.but i want a solution by chebyshev ineq for this problem.


search
number theory unsolved
number theory

Prove that exist infinite n,m :$ 3n^{2} \plus{}1\equal{}2m^{2} \plus{}2m$

This is the Pell equation .You can search on PEN to find an general problem with solution.

to TTspn :you are wrong ,it <=> $ (2m\plus{}1)^2\minus{}6n^2\equal{}3$

[quote="nguoigo"]to TTspn :you are wrong ,it <=> $ (2m \plus{} 1)^2 \minus{} 6n^2 \equal{} 3$[/quote]
The Pell's equation is :
$ ax^2\minus{}by^2\equal{}c$ with $ ab$ is not a square.


geometry
analytic geometry
trigonometry
ratio
similar triangles

Mandelbrot Round Four March 1991

8. The two circles at right have radii 1 and 3, and are internally tangent at $T$. What is the maximum area possible for a triangle with one vertex at $T$, another vertex on the small circle, and the third on the large circle?

the problem does include a diagram but I think the descriptions are pretty clear. If anyone has problems understanding I can provide a sketch, but for now, I won't.

one way would be to do a coordinate grid, and maximize the area using the cross product for area...

$\frac{1}{2}\left| \begin{tabular}{|ccc|} 0&0&1\\ \sin{\theta}&\cos{\theta}+1&1\\ 3\sin{\phi}&3\cos{\phi}+3&1\\ \end{tabular}\right|$

$\frac{3}{2}( \sin{\theta}*(\cos{\phi}+1)-\sin{\phi}*(\cos{\theta}+1))$

$\frac{3}{2}( \sin{\theta}*\cos{\phi}-\sin{\phi}*\cos{\theta}+\sin\theta-\sin\phi)$

$\frac{3}{2}(\sin{(\theta-\phi)}+\sin\theta+\sin-\phi)$

i don't know where to go from here...

I also used trig to start with, and just like Altheman, I was stuck after getting a trig expression to maximize.

[hide="Some idea..."]
Let $O$ be the center of the circle with radius $1$, and let $P$ be the center of the circle with radius $3$. $A$ is the vertex on circle $O$ and $B$ is the vertex on circle $P$. Let $\alpha=\angle ATO$ and $\beta=\angle BTO$.

Looking at a right triangle that contains $TA$ and the diameter of circle $O$ as its sides, we see that $TA=2\cos{\alpha}$. Similarly, $TB=6\cos{\beta}$.

We want to maximize the area of the triangle, which is given by
$\frac{1}{2}TA\cdot TB\cdot \sin{(\alpha +\beta)}$

$=\frac{1}{2}(2\cos{\alpha})(6\cos{\beta})(\sin{(\alpha +\beta)})$

$=6\sin{(\alpha +\beta)}\cos{\alpha}\cos{\beta}$

I don't know how to maximize the expression above, but I think this helps.
[/hide]

here is a geometric argument (i just read the geometry tidbits, i can't belive that it helps already)

[hide]let the triangle have verticies $T$ (point of tangency), $A$ where it intersects the smaller circle, and $B$ where it intersects the larger circle,

have $BT$ intersect the smaller circle at $X$

since the 2 internally tangent circles are homothetic in respect to $T$, and $X$ and $T$ are corresponding points, we have that $\frac{TX}{TB}=k$ for any point $T$

triangles $TAX$ and $TAB$ have the same height (to $TB$), and their bases are $TX$ and $TB$ respectively, so we have

$\frac{[TAX]}{[TAB]}=\frac{TX}{TB}=k$

therefore $[TAX]$ is directly proportional to $[TAB]$ so if we maximize $[TAX]$, then we maximize the area of $[TAB]$

we know that the maximum area for a triangle in a circle is just an equilateral triangle, so the area of that is $3*\frac{1}{2}*1*\sin{120}=\frac{3\sqrt{3}}{4}$

to find $k$, choose $B$ so $TB$ passes through the centers of the circles, so we find that $k=\frac{1}{3}$

so plugging this in to before, we have

$\frac{[TAX]}{[TAB]}=\frac{1}{3}=\frac{\frac{3\sqrt{3}}{4}}{[TAB]}$

$[TAB]=\boxed{\frac{9\sqrt{3}}{4}}$

also, i just noticed that the whole part about homotheity is unnecessary, you can make the same argument with similar triangles (just draw the line connecting the centers of the circles, and hit the smaller and larger circles at $C$ and $D$, respecitvely, then it is easy to show that $\triangle TAC \sim \triangle TBD$, thus the bases are in a constant ratio[/hide]

[quote="Altheman"]$\frac{3}{2}(\sin{(\theta-\phi)}+\sin\theta+\sin-\phi)$

i don't know where to go from here...[/quote]

[hide="Maximizing that expression"]
Rewrite $\phi \Rightarrow -\phi$ and you get $\frac{3}{2} \left( \sin(\theta + \phi) + \sin \theta + \sin \phi \right)$.  Intuitively, because the expression is symmetrical, the maximum value occurs at some $\theta = \phi$, so rewrite as $\frac{3}{2} \left( \sin (2\theta) + 2 \sin \theta \right) = \frac{3}{2} \left( 2 \sin \theta \cos \theta + 2 \sin \theta \right) = 3 \sin \theta (1 + \cos \theta)$.

Another bit of intuition gives the maximum value as $\theta = \frac{\pi}{3}, A = \boxed{\frac{9 \sqrt{3}}{4}}$, which agrees with the other answer given.[/hide]

[hide="Rigorously maximizing that expression"]
(To rigorously maximize the expression $\sin \theta (1 + \cos \theta)$:

$r = \sin \theta(1 + \cos \theta)$
$\frac{dr}{d\theta} = \cos \theta(1 + \cos \theta) + \sin \theta(- \sin \theta)$
$\frac{dr}{d\theta} = 2 \cos^2 \theta + \cos \theta - 1 = 0$
$u = \cos \theta \Rightarrow 2u^2 + u - 1 = 0$
$(2u - 1)(u + 1) = 0$
$u = \frac{1}{2}, -1$
$\theta = \frac{\pi}{3}, \frac{3\pi}{2}$

$\theta = \frac{3\pi}{2}$ gives the value of the expression as 0, so the other value of $\theta$ is the maximum.)[/hide]


geometry
pigeonhole principle
algebra unsolved
algebra

Consider a convex polygon with 2009 vertices. A scissor is used to cut along all of its diagonals, thereby dividing the original polygon into smaller convex polygons. How many vertices are there of a resulting polygon with the greatest number of edges ? :lol:

[quote="Differ"][hide]Look at a regular polygon. The desired polygon is exactly where you'd expect it to be.[/hide][/quote]

I don't understand . I was found a result that it's 2009  but I can't prove that

The result only holds for a regular polygon. It is easily verified that the middle polygon is, indeed, a 2009-edge polygon and thus is a 2009-vertex polygon. And if we have 2010 sides, by pigeonhole there must be three diagonals from one point.


There are two circles within one another as shown in the attachment { I'm horrible at paint , so pardon the figure. }.Anyway , the line segment you see in the figure is a chord to the larger circle and a tangent to the smaller circle.If the length of the line segment is 12 cm , and the line joining the centres of the two circles is $\|$ to the  chord, find the area of the larger circle - the area of te smaller circle.
I seriously dont understand this question... :blush:

By the way , the whole thing is one line {It looked like three separate line segments when I saw the figure...}

the drawing is very unclear... and what are those three lines that are drawn right above the inscribed circle?

From what is written, I don't think there is enough information to solve this.  Maybe there are things we are supposed to pick up from the diagram.  Do we know anything about smaller circle?  Is it also tangent to the larger circle?  Does it pass through the center of the larger circle?

Hey ,
Sorry for that unclear drawing , but then something went wrong when I uploaded....Anyways , those three line segments are well , pieces of a single line segment  , say , AB , which is a chord to the larger circle and a tangent to the smaller circle....{Join the three line segments --- it should hopefully be clear then........and I think the smaller circle touches the bigger circle at exactly one point....then how do you find the area of larger circle minus the area of the smaller circle ??

Maybe a better diagram will make the problem easier to understand.

$ \bigcirc A$ is tangent to $ \bigcirc B$.

If $a$ is the radius of $ \bigcirc A$ and $b$ is the radius of $ \bigcirc B$, then $a^2+6^2=b^2$. $b^2\pi-a^2\pi=\boxed{36\pi}$

Hey Nosoupforyou ,
thanks a real lot....for both the solution and that clear diagram...
  :lol:


geometry
geometric transformation
reflection
perpendicular bisector

$ 5.$ Let $ ABC$ be an acute-angled triangle with altitude $ AK$. Let $ H$ be its ortho-centre and $ O$ be its circum-centre. Suppose $ KOH$ is an acute-angled triangle and $ P$ its cicum-centre. Let $ Q$ be the reflection of $ P$ in the line $ HO$. Show that $ Q$ lies on the line joining the mid-points of $ AB$ and $ AC$.

Cool problem :lol:
point-reflect (half-turn) everything about midpoint $ N$ of $ OH$, the nine-point center. points lying on the nine-point circle will be mapped to diametrically opposite points. Hence the mid points of sides $ AB,AC$ maps respectively to midpoints of $ CH,BH$. the line joining them is the perpendicular bisector of $ HK$ as well, and that contains $ P$, the image of $ Q$ under the point-reflection. :)

Hey thanks Akashnil....you made my work simpler...
Due to a plethora of points I'll refer to...I've just indicated them in the diagram...(Note that I've drawn $ PS \perp HK$ and then extended it to $ R$ on $ BH$.)

Now What I wrote was that I first made a lemma stating that If $ R$ is the midpoint of $ BH$, and $ E$ is the midpoint of $ AC$, then $ RE$ is a diameter of the nine - point circle...and then proved it in the same style as PCM...

Now, coming back to the question....(I'm feeling too lazy so I'm just saying in brief....)
Now, since $ P$ is the circumcentre of $ \triangle OKH$,
Therefore, $ S$ is the midpoint of $ HK$, combining this with $ RS \parallel{} BK$ gives $ \boxed{R \ is  \ the \ Midpoint \ Of \ BH}$.
Now, since $ P$ is the circumcentre of $ \triangle OKH$, we also have $ N$ is the midpoint of $ OH \Rightarrow N$ is the nine - point centre.
So in $ \triangle s RNP, ENQ$ we have $ PN \equal{} NQ$(By reflection), $ EN \equal{} NR$(By lemma), $ \angle RNP \equal{} \angle QNE$(since $ \overline{RNE}$ is a straight line by lemma)
Therefore, $ \triangle RNP \equiv \triangle ENQ$ (By SAS)
and therefore, $ EQ \parallel{} RP \parallel{} BC \parallel{} DE$(First two we have proved by using the congruency, next two already proved, and next two follow from midpoint theorem)
So, $ EQ$ and $ DE$ coincide and thus the result follows..

Is it good, Akashnil?  :)

Can anyody or Ankush himself elaborate his so called easy solution written in a language too difficult to understand

[quote="ankush maths"]:lol: This is to all those who think this question is difficult.
Let the mid-point of AB be X and mid-point of AC be Y. Let Z be the foot of perpendicular from O to XY. Observe that triangle KOH is similar to triangle ZNO (Z,N,O correspond to K,O,H for the medial triangle), where N is the nine point centre. So, angle OZN = angle OQN = angle OKH . So, OZQN is a cyclic quad. So, angle QNO = angle QKO = 90 . So Q lies on XY.
 :lol:[/quote]

This is to all those who think that readers can comprehend such answers .....
[url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_HowWriteJustify.php]Just read this....[/url] (Solution Readers, NOT Mindreaders)

The official solns are out [url]http://olympiads.hbcse.tifr.res.in/uploads/inmo-2010[/url]

Well, for the medial triangle of triangle ABC ( I hope u know what this is, abhitayal ), Z is the foot of altitude on XY ( just like K is the foot of altitude on BC ), N is the circumcentre ( just like O is the circumcentre for triangle ABC ) and O is the orthocentre ( just like H is the orthocentre for triangle ABC ) . So, by SSS similarity triangles KOH and ZNO are similar ( I cant believe no one got this, it looked obvious to me ). So angle OZN = angle HKO . And since Q is the reflection of circumcentre P of triangle KOH, angle OQN = angle HKO . ( This is just like the fact that if O' is the image of O in BC , angle BO'D = angle A, where D is the mid-point of BC as here Q is the image of P and N is the mid-point of OH ). Now since angle OZN = angle OQN , OZQN is a cyclic quad ( angles in the same segment are equal, abhitayal ). Now, angle OZQ + angle ONQ = 180 and since angle ONQ = 90 , angle OZQ = 90 . Hence, Q lies on XY as angle OZQ = angle OZX = 90.
Are u satisfied now, abhitayal ? Or should I give the proof of the fact that in a cyclic quad , sum of opposite angles is 180 ?

I hope the medial triangle is the triangle formed by joining the midpoints of the sides of triangle ABC(ie generally D,E,F or in your case X,Y and another point which is not Z) :rotfl:
Secondly, is this quite obvious or an assumption that the cyclic quad formed is  O[b]ZQ[/b]N and not O[b]QZ[/b]N :?:
Lastly, I wouldn't mind you giving me a proof of the fact that the sum of oppossite angles in a cyclic quad is 180 degrees. My proof of this fact is quite cumbersome. :rotfl:

[hide="PS"]Thanks for using my name a number of times in your post. Made me feel honoured :rotfl: [/hide]

Is Ankush maths - Ankush Das INMO topper 2009?

@Aravind
Yes!! Ankush maths is Ankush Das

Whether the cyclic quad is OZQN or OQZN , my proof will work in both the cases. For the proofs, refer to class IX or X R.D. Sharma. It contains a rigorous proof. :rotfl:

Well the exact same method will not be applicable if quad is OQZN but never mind :P
As for that proof I already know the RD proof
thought that the [i]INMO topper[/i] would provide me with a better one :rotfl:


trigonometry
function
inequalities
inequalities unsolved

Prove that for $k>10$ Nazar may replace in the following product some one $\cos$ by $\sin$ so that the new function $f_{1}(x)$ would satisfy inequality $|f_{1}(x)|\le 3\cdot 2^{-1-k}$ for all real $x$.
\[f(x) = \cos x \cos 2x \cos 3x \dots \cos 2^{k}x \]
[i]N. Agakhanov[/i]

Maybe it should be $ k>1$ instead of $ k>10$, because
this holds for all $ k>1$, I think. Its a bit strange to write $ k>10$.
Could you please check original text?

Hmmm... The text seems correct (it says $ k>10$ in my copy of ARMO 2007 problems...)

So whats wrong here: if k>1 then we can replace cos3x with
sin3x and we have
$ |f_{1}(x)|\leq|\sin3x\prod_{i=0}^{k}\cos2^{i}x|= |(3\sin x-4\sin^{3}x)\frac{\sin2^{k+1}x}{2^{k+1}\sin x}|= |(3-4\sin^{2}x)\frac{\sin2^{k+1}x}{2^{k+1}}|\leq\frac{3}{2^{k+1}}???$

[color=#FF0000]Moderator Edit: do not quote the entire post above you.[/color]

It seems correctly,
but then it is for $k>10$ a bit trying with $3?$

It seems $f(x) = \cos x \cos 2x \cos 3x \cdots \cos 2^k x$, which seems to indicate $f(x) = \prod_{j=1}^{2^k} \cos jx$, has suddenly transmogrified into $f(x) = \cos 3x \prod_{i=0}^{k} \cos 2^i x$  :) EDIT. [color=#0000FF]Ah, I just noticed the idea below had already been incorporated, by having written $|f_1(x)| \leq \cdots$; sorry.[/color]

However, since $|\cos jx| \leq 1$ for all $j$, we indeed have $|f(x)| \leq |\cos 3x|\cdot \left | \prod_{i=0}^{k} \cos 2^i x \right |$, and now we can replace $\cos 3x$ with $|\sin 3x|$ and continue as above. So the trick is to keep just the factors containing powers of $2$ (in order to be able to handle their product).


How many different two-letter monograms are possible if letter may be repeated?

I need help on the problem. :?:

im guessing 26^2=676 but idk what a monogram is

a monogram is a monogram is a motif made by overlapping or combining two or more letters or other graphemes to form one symbol.

But either Way, it is 26x26=676

[u]26 [/u]         possibilities
   1st Letter

 [u]26[/u]           possibilities
2nd Letter

26possibilities  X  26possibilities = 676 possibilities

Get it?
Great!
[/u]

Well, there are $ 26$ choices for each letter

(If you don't know that there are $ 26$ letters in the alphabet,

forget it...)

so $ 26^2\equal{}676$.


combinatorics proposed
combinatorics
graph theory
college contests

In a connected simple graph every vertex has a degree at least 3.
Prove that the graph contains a cycle such that the graph remains
connected when the edges of this cycle are deleted.

Ok, after some days thinking about it, I got it  :)

The graph has to be finite, otherwise the result does not hold (consider an infinite tree with a root with three sons and each other vertex has one father and two sons).

Let's start from any vertex, and label it $ 1$.
Then, for each $ i$, label $ i\plus{}1$ all the vertices which are not already labelled and are adjacent to at least one vertex with label $ i$.

Note that :
a) since the graph is finite, the procedure will eventually end. Let $ p$ be the maximum used label.
b) for each $ i$, the subgraph whose vertices are those with label not greater than $ i$ is connected (any two vertices are connected to the vertex with label $ 1$).

Now, consider the subgraph whose set of vertices are all vertices with label $ p$. If this graph contains a cycle then stop.
Otherwise, extend the subgraph by adding the vertices with label $ p\minus{}1$ (and the corresponding edges). If this graph contains a cycle then stop. Otherwise, add the vertices with label $ p\minus{}2$ and so on.
This new procedure will eventually stop because, the whole graph contains a cycle (it has $ n$ vertices and $ \frac {3n}{2} \geq n$ edges).
Let $ k$ be the maximal integer such that there is a cycle in the subgraph whose set of vertices is those with labels at least $ k$.
Among these cycles, choose one, say $ C$, with minimal length.
I claim that $ C$ can be deleted without destroying the connectivity of the whole graph.

Assume, for a contradiction, that deleting $ C$ detroys the connectivity. Then, from b), there is a connected component, say $ S$, which contains only vertices with label at least $ k\plus{}1$ (each vertex with label $ k$ in $ C$ is adjacent to a vertex with label $ k\minus{}1$ and the corresponding edge has not been deleted since $ C$ does not contain any vertex with label less than $ k$).
Moreover, the maximality of $ k$ forces $ S$ to contain no cycle.

Among the vertices in $ S$, at least one belongs to the cycle $ C$, otherwise deleting $ C$ would not have destroy the connectivity of $ S$ with the initial vertex with label $ 1$.

Let $ A$ be a vertex in $ S \cap C$. Exactly, two of the edges with endpoint $ A$ have been deleted, so it remains one, say $ A\minus{}X_A$. Then :
- $ X_A \notin C$ :
Assume, for a contradiction, that $ X_A \in C$. Since $ A\minus{}X_A$ is not deleted, it follows that $ X_A$ is not adjacent to $ A$ in $ C$. But, then, using the edge $ A\minus{}X_A$ we will find a cycle with vertices with labels no less than $ k$ and with length less than the length of $ C$, which contradicts the minimality of $ C$.
- $ X_A \in S$ thus, from above, $ X_A$ has label at least $ k\plus{}1$.

1) If there are two vertices, say $ A$ and $ B$, in $ S \cap C$ such that $ X_A \equal{} X_B$ :
Let $ X$ denotes this common neighbour. The minimality of $ C$ forces $ C$ to be triangle $ ABP$ or a quadrilateral $ APBQ$ (otherwise, using the edges $ A\minus{}X\minus{}B$ we would find a cycle with length less than $ C$).
- If $ C \equal{} ABP$ then the cycle $ ABX$ contradicts the maximality of $ k$.
- If $ C \equal{} APBQ$ then $ P$ and $ Q$ have label $ k$ (otherwise $ APBX$ or $ AQBX$ would contradicts the maximality of $ k$). It follows that neither $ P$ nor $ Q$ belongs to $ S$, and that the only vertices in $ C \cap S$ are $ A$ and $ B$.
Let $ m \equal{} |S|$. Since $ X$ and its three neighbours belong to $ S$, we have $ m \geq 4$.
Now, each vertex in $ S$ has degree three except $ A$ and $ B$ which have degree $ 1$. The sum of the degrees in $ S$ is $ 3(m\minus{}2)\plus{}2 \equal{} 2m \plus{} (m\minus{}4) \geq 2m$. Therefore, the number of edges in $ S$ is at least $ m$, which forces $ S$ to contain a cycle. Contradiction.

2) If no two vertices in $ C \cap S$ share the same neighbour in $ S$.
Let $ m \equal{}|S|$. Let $ A_1, \cdots, A_s$ be the vertices in $ S \cap C$, and $ X_1, \cdots , X_s$ be their respective neighbours in $ S$. From above, these $ 2s$ vertices are pairwise distinct, so that $ m \geq 2s$.
As above, each $ A_i$ has degree $ 1$ in $ S$, and each other vertex in $ S$ has degree $ 3$.
Thus, the sum of the degrees in $ S$ is $ 3(m\minus{}s)\plus{}s \equal{} 2m \plus{} (m\minus{}2s) \geq 2m$, which leads to the same contradition as above.

Then, in each case, we reach a contradiction. It follows that $ C$ can be deleted without destrotying the connectivity of the whole graph, and we are done.

Pierre.

Just a comment on the above solution, the way I see it:

We construct a spanning tree by breadth-first-search. Such tree has the following property. All edges, except that of the tree, are cross edges regarding the tree. Moreover these cross edges are between vertices of the same or neighbouring levels.

Now, we number the vertices with their level, the root being at level $1$, and initially consider the set of vertices with the highest level. Then we begin to expand down this set by including consecutively the lower levels, one by one, till a cycle(one or many) appears at the first time. We remove the edges of the shortest cycle (if they are more than 2)- call it $C$- and let's prove the remaining graph is still connected.

Suppose we have dropped down to the level $k$ (including it). All vertices of the tree up to $k$ (including it) are still connected. Let $v$ be a vertex of $C$ with level higher than $k$. Since its degree is $3$ or more, there is a edge $v v_1$ with $v_1\not\in C$. There are two possibilities: $v_1$ is one level upper than $v$, lower, or $vv_1$ is a cross edge.

In the first case we go from $v$ to a top leaf (we can not meet other edges of $C$ (because it would make a higher cycle or a cycle with small length- a little argument needed here), then from that leaf- say $v_2$ there must be a cross edge-say $v_2 v_3$. Going down from $v_3$ and $v_2$, let $v_4$ be the the first vertex they meet. It can be easily seen (by the same argument) there are not vertices of $C$ between $v_4, v_3$ and $v_4, v_2$ . The level of $v_4$ is less or equal to $k$, because otherwise there would be a cycle higher than $C$. It means $v$ is still connected to the root via: $v v_1 v_2 v_3 v_4- \text{ root}$.
The other two cases are similar.


Solve the equation $ x^2\minus{}2x\minus{}2007y^2\equal{}0$ in positive integers.

Let $ z \equal{} x\minus{}1 \implies z^2 \equal{} x^2 \minus{} 2x \plus{} 1 \implies z^2 \minus{} 1 \equal{} x^2 \minus{} 2x$

Then, we have $ z^2 \minus{} 2007 y^2\equal{} 1$, a [url=http://en.wikipedia.org/wiki/Pell_equation]Pell equation[/url].
[hide="Solving a Pell Equation"]We first look at the convergents of the [url=http://en.wikipedia.org/wiki/Continued_fraction]continued fraction[/url] of $ \sqrt{2007}$.

The first few are $ \left \{ 44, 45, \frac{179}{4}, \frac{224}{5}, \ldots \right \}$

We then try out the numerators and denominators of these as $ z$ and $ y$ values, respectively, until we find a solution.

$ (z,y) \equal{} (44,1), \; (45, 1), \; (179,4)$ don't work, but $ (224,5)$ works.[/hide]
The fundamental solution is $ z_1 \equal{} 224, \; y_1 \equal{} 5$

Subsequent solutions are found using the recurrence relations:
$ z_{i\plus{}1} \equal{} z_1 z_i \plus{} n y_1 y_i, \qquad y_{i\plus{}1} \equal{} z_1 y_i \plus{} y_1 z_i$
which in this case becomes:
$ z_{i\plus{}1} \equal{} 224 z_i \plus{} 10035 y_i, \qquad y_{i\plus{}1} \equal{} 225 y_i \plus{} 5 z_i$

Equivalently, $ \boxed{(x_1, y_1) \equal{} (225,5)}$, and we have the recurrence relation:
$ \boxed{x_{i\plus{}1} \equal{} 224 x_i \plus{} 10035 y_i \minus{}223, \qquad y_{i\plus{}1} \equal{} 224 y_i \plus{} 5 (x_i\minus{} 1)}$


Find the equation of the circle passing through the point $(-3,1)$ and containing the points of intersection of the circles $x^2+y^2+5x=1$ and $x^2+y^2+y=7$

The desired equation of the circle can be written $(x^2+y^2+5x-1)+k(x^2+y^2+y-7)=0$ for arbiterary real numbers $k.$ Since this circle passes through the point $(-3,1),$ we have $-6+4k=0\Longleftrightarrow k=\frac{3}{2}.$
Thus the answer is $5x^2+5y^2+10x+3y-23=0.$

[quote="kunny"]The desired equation of the circle can be written $(x^2+y^2+5x-1)+k(x^2+y^2+y-7)=0$ for arbiterary real numbers $k.$ Since this circle passes through the point $(-3,1),$ we have $-4+4k=0\Longleftrightarrow k=1.$
Thus the answer is $2x^2+2y^2+5x+y-8=0.$[/quote]

kunny ,I think there is a typo there because  $k=\frac{3}{2}$   :)

You are right. :blush: I will edit.

kunny


trigonometry
inequalities unsolved
inequalities

Prove or disprove
$ a+b+c\geq 3\sqrt{3}R $
a,b,c sides of a triangle

Silouan let me help!!

Using sine law the ineq in question is equivalent to
$\displaystyle  sinA+sinB+sinC\leq \frac{3\sqrt{3}}{2}$
which holds due to Jensen's ineq.

So the $\geq$ sign should actually be a $\leq$ sign! ok?? :)

See this for all such trignometry results...

In fact, this is Mitrinovic's inequality. ;)

$r_a,r_b,r_c$ are the roots of the ecuation: $x^3-(4R+r)\cdot x^2+s^2\cdot x-rs^2=0$.
$S=sr=\sqrt {s(s-a)(s-b)(s-c)},\ abc=4RS$.
*****************************************************************
$1^{\circ}.\ 3(r_a\cdot r_b+r_b\cdot r_c+r_c\cdot r_a)\le (r_a+r_b+r_c)^2\Longrightarrow\underline{\overline {| \ s\sqrt 3\le 4R+r\ | }}$

$2^{\circ}.\ a^2\ge a^2-(b-c)^2,\ a.s.o.\Longrightarrow a^2\ge 4(s-b)(s-c)\ a.s.o.\Longrightarrow$

${3^{\circ}.\ abc\ge 8(s-a)(s-b)(s-c)\Longrightarrow \underline {\overline | \ R\ge 2r\ | }}$

$4^{\circ}.\ \frac{(s-a)+(s-b)+(s-c)}{3}\ge \sqrt [3] {(s-a)(s-b)(s-c)}\Longrightarrow \underline {\overline {| \ 3r\sqrt 3\le s\ | }}$

At once we add the relation (1) with the relation (4) easily altered $r\le \frac{s}{3\sqrt 3}$
and obtain the relation$:\ s\sqrt 3\le 4R+\frac{s}{3\sqrt 3}\Longrightarrow \underline {\overline {| \ s\le \frac{3\sqrt 3}{2}\cdot R\ | }}$
****************************************************************
$\blacksquare\clubsuit\ \underline {\overline {| \ 3\sqrt 3 \cdot r\le s\le \frac {3\sqrt 3}{2}\cdot R\ ;\ \ \sqrt 3\cdot s\le 4R+r\ | }}\ \clubsuit\blacksquare$

****************************************************************


calculus
derivative
integration
calculus computations

:oops:
hehe
i searched for "limits and derivatives" on google and it says that it's differential calculus
can anyone explain in simple terms to a nub (me)?
thx to anyone who will
 :ninja:

Well, briefly the base of the calculus is the “limit”. First a deep study becomes on the limits thus to see “derivatives” and the the opposite process “integration”. Knowing these things later others like differential equations, multiple integration among others.

haha thanks i found out how though
thx for trying


if $x_1=\frac{1+\sqrt{5}}{2}$ and $x_2=\frac{1-\sqrt{5}}{2}$
prove that $(x_1)^n-(x_2)^2$ is an integer number. :D

[quote="dondigo"]Didn't you mean $x_1^n+x_2^n$ if yes, then:
$x_1+x_2$ and $x_1x_2$ are integers.
$x_1=a$
$x_2=b$

$a^n+b^n=(a^{n-1}+b^{n-1})(a+b)-ab(a^{n-2}+b^{n-2})$
It's ease to prove it from induction.[/quote]
no i mean exactly $x_1^n-x_2^n$,sorry.(but your solution for that is nice.) :D

yes ,sorry, you are right, it is :
$\frac{x_1^n-x_2^n}{\sqrt{5}}$

Firstly: Jensen, please edit the topic, because there is still an error. Secondly I've done it :)
We have to prove, that: $\frac{(x_1)^n-(x_2)^n}{\sqrt{5}}$
$\frac{1}{\sqrt{5}}\cdot ((\frac{1+\sqrt{5}}{2})^n-(\frac{1-\sqrt{5}}{2})^n))$

Notice, that this sequence is a Fibbonaci sequence, so it must be integer :D

look at the binomial expansions of those, and you will notice that the even exponents for $\sqrt{5}$, cancel out because they are both positive, and then they are subtracted, in contrast, the odd exponents do not cancel out, and leave a $\sqrt{5}$ which is later divided out leaving an integer


The base of a triangle is twice as long as a side of a square and their areas are the same.  Then the ratio of the altitude of the triangle to the side of the square is:

$ \textbf{(A)}\ \frac{1}{4} \qquad
\textbf{(B)}\ \frac{1}{2} \qquad
\textbf{(C)}\ 1 \qquad
\textbf{(D)}\ 2 \qquad
\textbf{(E)}\  4$

[hide="Click for solution"]
Let $ s$ be the side of the square and $ h$ the altitude of the triangle.  Then $ \frac{1}{2}h \cdot 2s\equal{}s^2 \implies h\equal{}s$, so $ \frac{h}{s}\equal{}1$, or $ \boxed{\textbf{(C)}}$.
[/hide]


The New York Math Circle is a new math circle for students and teachers in the New York City area.  Its website is [url]http://www.nymathcircle.org[/url].

Classes for students meet on Saturday mornings from 11 am to 1 pm at NYU.  The first class is on March 1.  The class for 6th through 8th graders is taught by Joy Hsiao.  The class for 9th through 11th graders is taught by David Hankin.  The cost  for this semester is 105 dollars.  If you are interested, the organizers would like to receive your registration by February 18.

The class for teachers meets on Wednesday evenings from 5 pm to 7 pm.  The first class is on February 27.  The class is taught by Larry Zimmerman.  The cost for this semester is 85 dollars.

Check it out!

For those who don't know them, David Hankin and Larry Zimmerman are the retired coaches of the Hunter and Brooklyn Tech math teams, respectively, and two of the best teachers you're likely to find anywhere.  David has been the chair of the AIME committee for many years, and Larry was a long-time writer of the ARML and NYSML contests.

Update: The class for 6th through 8th graders has been canceled for this semester.  It is now planned for Fall 2008.

The class for 9th through 11th graders (taught by Mr. Hankin) is still on.  The class for teachers (taught by Mr. Zimmerman) is still on.  Each class meets for the first time this week.  Each has a few spots available as of today.


function
complex numbers
real analysis
real analysis unsolved

Does there exist a differentiable function $ f: \mathbb{R} \to \mathbb{R}$, not identically zero, such that $ f'(x) \equal{} f(x\plus{}1)$ for all $ x$?

There is, as I recall, a really long and classic thread about this.  The basic idea is to take $ f(x) \equal{} e^{cx}$ for some $ c \in \mathbb{C}$ such that $ c \equal{} e^c$.  The original thread went on to prove that such numbers existed and there was a lot of discussion about generalizations, as I recall.  Then the real part of $ f$ is such a function.

I like "classic" as the descriptor.  Evidently, I was responsible for the thread you're thinking of:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=8145
The question posted here has also appeared elsewhere:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=218458
And here's another related problem:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=19406
(I found these threads by searching for posts by user Ravi B that contained the word "complex.")


A jar contains 28 marbles. Half of the marbles are red. Half of the non-red marbles are white and the rest are blue. Todd chose a white marble at random and kept it. What is the probability that when Hosea now draws a marble it will also be white? Express your answer as a common fraction.

From the given, there are $ 14$ red marbles, $ 7$ white marbles, and $ 7$ blue marbles. When one was chosen, that left $ 7 \minus{} 1 \equal{} 6$ white marbles. Therefore, the probability is $ \frac {6}{27} \equal{} \boxed{\frac29}$.


1. Find all functions $f: \mathbb R \to \mathbb R$ such that \[ f(x^2\minus{}y^2)\equal{}(x\minus{}y)(f(x)\plus{}f(y))\] holds for all real values of $x$ and $y$.

2. Find all functions $f: \mathbb R \to \mathbb R$ such that \[ xf(x)\minus{}yf(y)\equal{}(x\minus{}y)f(x\plus{}y)\] for all $x,y \in \mathbb R$.

[quote="Johan Gunardi"]
2. Find all functions on real such that $ xf(x) \minus{} yf(y) \equal{} (x \minus{} y)f(x \plus{} y)$.[/quote]

Denote $ g: R \rightarrow R$ satisying $ g(x) \equal{} f(x) \minus{} f(0)$. Thus, $ g(0) \equal{} 0$ and
$ xg(x) \minus{} yg(y) \equal{} (x \minus{} y)g(x \plus{} y) (1)$

In (1), putting $ y \equal{} \minus{} x$ yields $ x(g(x) \plus{} g( \minus{} x)) \equal{} 0 \rightarrow \minus{} g(x) \equal{} g( \minus{} x)$
In (1), putting $ y : \equal{} \minus{} y$ yields $ xg(x) \minus{} ( \minus{} y)g( \minus{} y) \equal{} (x \plus{} y)g(x \minus{} y)$, implying $ xg(x) \minus{} yg(y) \equal{} (x \plus{} y)g(x \minus{} y) (2)$

Combining (1) and (2), we obtain $ (x \minus{} y)g(x \plus{} y) \equal{} (x \plus{} y)g(x \minus{} y)$, or $ \frac {g(x \plus{} y)}{x \plus{} y} \equal{} \frac {g(x \minus{} y)}{x \minus{} y}$

Setting $ x \equal{} \frac {u \plus{} v}{2}, v \equal{} \frac {u \minus{} v}{2}$, we obtain $ \frac {g(u)}{u} \equal{} \frac {g(v)}{v}$ for all non-zero real number u,v. So $ g(x) \equal{} ax$ for some constant a.

Hence the result is $ f(x) \equal{} ax \plus{} b$

[quote="Johan Gunardi"]1. Find all functions on real such that $ f(x^2 \minus{} y^2) \equal{} (x \minus{} y)(f(x) \plus{} f(y))$.[/quote]
The solution to the first is the same than ll931110 gave for the second :

Let $ P(x,y)$ be the assertion $ f(x^2 \minus{} y^2) \equal{} (x \minus{} y)(f(x) \plus{} f(y))$

$ P(x,x)$ $ \implies$ $ f(0)\equal{}0$
$ P(x,0)$ $ \implies$ $ f(x^2)\equal{}xf(x)$
$ P(\minus{}x,0)$ $ \implies$ $ f(x^2)\equal{}\minus{}xf(\minus{}x)$ and so $ f(\minus{}x)\equal{}\minus{}f(x)$ $ \forall x$
$ P(x,\minus{}y)$ $ \implies$ $ f(x^2\minus{}y^2)\equal{}(x\plus{}y)(f(x)\minus{}f(y))$

and so $ (x \minus{} y)(f(x) \plus{} f(y))\equal{}(x\plus{}y)(f(x)\minus{}f(y))$
and so $ 2xf(y)\equal{}2yf(x)$

And so the solution $ \boxed{f(x)\equal{}ax}$ (after easily checking that this is indeed a solution)


I'm having a little trouble with the following:

$ \int_{-\frac{\pi}{2}}^\frac{\pi}{2}\cos{2\theta}\ d\theta = \left. \frac{1}{2}\sin{2\theta}\right|_{-\frac{\pi}{2}}^\frac{\pi}{2}$

As you can see, at the evaluation line thingy, the $ \pi/2$ is kind of mis-aligned... Is it  possible to move it back a little so it lines up with $ -\pi/2$??

[quote="pkerichang"]$ \int_{-\frac{\pi}{2}}^\frac{\pi}{2}\cos{2\theta}\ d\theta = \left. \frac{1}{2}\sin{2\theta}\right|_{-\frac{\pi}{2}}^\frac{\pi}{2}$[/quote]You need to put in an invisible minus sign before the upper limit, and you do this using \phantom{-}  $ \int_{-\frac{\pi}{2}}^\frac{\pi}{2}\cos{2\theta}\ d\theta = \left. \frac{1}{2}\sin{2\theta}\right|_{-\frac{\pi}{2}}^{\phantom{-}\frac{\pi}{2}}$
You could instead put in + to get $ \int_{-\frac{\pi}{2}}^{+\frac{\pi}{2}}\cos{2\theta}\ d\theta = \left. \frac{1}{2}\sin{2\theta}\right|_{-\frac{\pi}{2}}^{+\frac{\pi}{2}}$ which I think looks nicer.


geometry
3D geometry
calculus
calculus computations

\[ y^2 \equal{} \frac {x} {{x \plus{} y}}\]
dy/dx?

I can't get it
halp

lol,I know.
But I put it in wolfram alpha and I get another result.
thanks

I get the answer as $ \frac {dy}{dx}  \equal{} \frac{y}{ ( x\plus{}y)^{2} 2y \plus{} x }$

So, what does this "other result" look like? Does it have a bunch of square and cube roots? If so, that's coming from solving the cubic equation for $ y$. That's possible, but definitely not worthwhile for most applications.

This is why people should stop plugging into Mathematica for quick answers.  Try solving it by hand first.  And besides, if you can solve it by hand and then show that it's the thing as what Mathematica spits out, you learn something in the process!

$ y^2\equal{}\frac{x}{x\plus{}y}$ $ \Rightarrow$ $ x\equal{}\frac{y^3}{1\minus{}y^2}$
so $ dx/dy\equal{}\frac{y^2(y^2\minus{}3)}{(1\minus{}y^2)^2}$
Then we can get $ dy/dx$


Given is a two-player-game without random inpact. Does the statement hold in every game.

(i)  There exist a strategy for one player with he can't loose.

( random inpact means an inpact wich no player can influence it.)

not necesarily; for instance chess would count as a game with no random impact (Go is another example) but i would v much doubt if there is a winning strategy even before the start of the game.
what however is true is that if the game is guaranteed to end in some finite number of steps(absolute bound on the duration of the game) and if the game definitely produces a winner/loser(i.e. there cannot be draws) then the statement is true. it is an easy consequence of Konig's Infinity lemma.

Actually, I think that we could consider that theoretically, in chess, players do have a strategy! The game has a finite number of states, so either each player can guarantee a draw, or he can win....so He doesn't lose!!!
Ok, I think that in this problem, we should be concerned about the number of states of the game, not moves.....
Maybe I am wrong...but this is just an opinion..... :|

well even if chess can end in finitely many steps there is no upper bound on the number of moves the game lasts..
also when i say there is a strategy i mean that even before the start of a game by playing correctly one or the other player should win(or ensure he/she doesn't lose). that is not clear at all with chess. in fact some people believe that chess is more favorable to the first player, but... :)

Okay, I have to repear what I said, but chess has a FINITE number of STATES........Basicly, a state is a position of the figures on the table + who should move at that moment! We are actually not interested in the number of moves that the players made before that.....
Now, to settle things down, consider the directed graph of the states of the game chess....
(We shall consider only the connected component of the states achievable from the initial position)
Now, we color it's vertices according to the following rule:
1. Mark all possition that are obviously losing for the player at move with the black color!
2. For any vertex that is connected only with black vertices, mark it white!
3. For any vertex that is connected only with white vertices, mark it black!
Repeat these steps until you can't do anything else!
Of course, most probably, we will not mark the initial position with any color....(I am not sure....If I knew the answer, people wouldn't play chess :D ). Now, ALL the unmarked positions are actually DRAW positions....Any of the players can ensure that the other doesn't win!
So, I think this shows that chess is an uninteresting game! :D
Actually, I believe the number of states in chess is bigger than our most optimistic expectations of a computer's possibilities in many many years :)

I believe it states that an infinite connected graph in which every vertex has finite degree has an infinite path.


inequalities
inequalities proposed
algebra

If $a$, $b$, $c$ are the sidelenghts of a triangle, show that \[\left|\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}\right|<\frac{1}{16}\]

[hide]We have $|\sum\frac{a-b}{a+b}|=|\frac{(a-b)(b-c)(c-a)}{(a+b)(b+c)(c+a)}|$. Written $a=x+y,b=y+z,c=z+x$, here $x,y,z$ are positive reals. Now WLOG assume that $1=z<x<y$, and I used real analysis. :)[/hide]

[quote="Jutaro"]By the way, the right-hand side of the inequality is known not to be the best possible. I found out in Engel's book (problem 7.58) that the smallest upper bound is $\frac{8\sqrt{2}-5\sqrt{5}}{3}$. Could you show me how to prove this using analysis?[/quote]
Okie, here is my solution
Let $a=y+z,b=z+x,c=x+y$ then
\[LHS=\left\vert\\\frac{(x-y)(y-z)(z-x)}{(2x+y+z)(2y+z+x)(2z+x+y)}\right\vert\\ \]
Suppose $x \ge ge z$, setting $m=x-z,n-y-z$ then $m \ge n \ge 0$, we have
\[LHS=\frac{(m-n)mn}{(2m+n+3z)(2n+m+3z)(m+n+3z)}\le \frac{(mn(m-n)}{(2m+n)(2n+m)(m+n)}\]
Setting $m=tn$ $(t \ge 1)$, It suffices to determine the maximum value of
\[f(t)=\frac{t(t-1)}{(t+1)(2t+1)(t+2)}\]
But it's easy. :)


Any calculus solution (I know a probability solution)? This appeared on an integral inequality handout.


$a_1,a_2,\cdots,a_n$ are all positive integers,such that $1 \leq a_{i+1}-a_i \leq 2$
can $\sum_{i=1}^n \frac{ 1}{a_i}$ be intergers?

come on.i think it is hard(at last to me),i haven't sloved it.


let[color=blue] $ ABCD$ [/color]an insciptible quadrilateral such that [color=blue]$ AB>CD$[/color] and [color=blue]$ BC<AD$[/color]. let  $ P$  a point in [$ AB$] such that[color=blue] $ AP\equal{}CD$[/color] and $ Q$ a point in [$ AD$] such that [color=blue]$ AQ\equal{}BC$[/color]
the right  $ (AC)$ intersect $ (PQ)$ in $ M$.
prove that $ M$ is the  middle of [color=black] [$ PQ$][/color]

Easy and nice problem !
[quote="mr.danh"]ABCD is an cyclic quadrilateral such that AB > CD and BC < AD. let  P  a point in AB such that AP = CD and Q a point in [AD] such that AQ = BC. The line AC intersect PQ in E. Prove that E is the middle of [PQ][/quote]
$ I\equal{}AC\cap BD$. Let M,N on the line AC such that $ DM\parallel BN\parallel PQ$.
$ \frac{DM}{QE}\equal{}\frac{AD}{AQ}\equal{}\frac{AD}{BC}\equal{}\frac{IA}{IB}\Rightarrow QE.IA\equal{}DM.IB$
Similarly, $ PE.IA\equal{}BN.ID$
$ \frac{IB}{ID}\equal{}\frac{BN}{DM}$ (because $ DM\parallel BN$), then $ DM.IP\equal{}BN.ID$
So, QE=PE.


search
induction
number theory unsolved
number theory

[quote="farshid_lk_park"]prove that :
A = 1/a + 1/a+1 + ... + 1/a+k
 is not a natural number[/quote]
a ?????????????? complete the exercice (the condition present 1/2 of the solution  :whistling:

Obviously, if $ a\equal{}1$, we can search for integers of the form $ \frac{1}{2}\plus{}...\plus{}\frac{1}{k\plus{}1}$. Thus, we can assume $ a>1$, $ k\geq 1$ (unless we treat $ a\equal{}1$, $ k\equal{}0$ as a solution, which is the only solution when $ k\equal{}0$ since $ \frac{1}{a}<1$ for $ a>1$)

Let $ p$ denote the largest prime such that $ p|a\plus{}i$ for some $ 0\leq i\leq k$. By Bertrand's postulate, we know there exists a prime $ p$ such that $ n<p<2n$ for every $ n>1$. So, if $ k\geq a$, there exists a prime number $ p\equal{}a\plus{}s$ ($ s$ is a natural number less or equal to $ k$). Let $ A$ be a natural number such that $ Aa(a\plus{}1)...(a\plus{}k)\equal{}(a\plus{}1)...(a\plus{}k)\plus{}a(a\plus{}2)...(a\plus{}k)\plus{}...\plus{}a(a\plus{}1)...(a\plus{}k\minus{}1)$. Now, all the products on the right side except $ a(a\plus{}1)...(a\plus{}s\minus{}1)(a\plus{}s\plus{}1)...(a\plus{}k)$ will be divisible by $ p$. Also, the left side will be divisible by $ p$. So, $ a(a\plus{}1)...(a\plus{}s\minus{}1)(a\plus{}s\plus{}1)...(a\plus{}k)$ must be divisible by $ p$. As $ p$ is not present in this product, $ 2p$ must obviously be present in the product. So, $ a\plus{}k\geq 2a\plus{}2s$ and we have $ A\equal{}\frac{1}{a}\plus{}\frac{1}{a\plus{}1}\plus{}...\plus{}\frac{1}{a\plus{}s}\plus{}...\plus{}\frac{1}{2a\plus{}2s}\plus{}...\plus{}\frac{1}{a\plus{}k}$. However, then there, again, by Bertrand we know that there exists a prime between $ a\plus{}s$ and $ 2a\plus{}2s$. So, $ p$ is not the largest such prime. That's a contradiction and we conclude that $ k<a$.
So, we have $ A\equal{}\frac{1}{a}\plus{}...\plus{}\frac{1}{a\plus{}k}<\frac{1}{a}\plus{}...\plus{}\frac{1}{2a}$. We can use induction in proving that $ A<\frac{1}{a}\plus{}...\plus{}\frac{1}{2a}<1$ for $ a>2$. ($ \frac{1}{a}>\frac{1}{2a\plus{}1}\plus{}\frac{1}{2a\plus{}2}$ is equivalent to $ 4a^2\plus{}6a\plus{}2>2a^2\plus{}2a\plus{}2a^2\plus{}a$, which is equivalent to $ 3a\plus{}2>0$, which, of course, holds.) So, $ a\equal{}2$. Now, obviously $ k\equal{}1$ or $ k\equal{}2$. However, $ \frac{1}{2}\plus{}\frac{1}{3}$ is not natural. We arrive at the conclusion that A can not be a natural number. Q.E.D.


Can you prove this for all real $ a, b, c$ (even negative values can occur):

$ \sum_{cyc}a^4b^2 \plus{} 2\sum_{cyc}a^3bc^2 \plus{} 9a^2b^2c^2 \ge 6\sum_{cyc}a^3b^2c$

[quote="Zuza"]Can you prove this for all real $ a, b, c$ (even negative values can occur):

$ \sum_{cyc}a^4b^2 + 2\sum_{cyc}a^3bc^2 + 9a^2b^2c^2 \ge 6\sum_{cyc}a^3b^2c$[/quote]
It's $ (a^2b+b^2c+c^2a-3abc)^2\geq0.$ :wink:


[quote="Rep123max"]Remember that question from chapter with the sum of all even five digit numbers that can be made using the digits 1, 3, 4, 5 and 9 only once? What if it could be repeated or if it was 4 digits numbers, or 4 digits numbers that can repeat. Basically, first, how would you tackle it if repetition is allowed?[/quote]
There are 625 ways this number could be made. 11114+33334+44444+55554+99994. So the average would be 48888.  Multiply by 256.  which would be 30555000.  I think.

edit.  Miscounted ways.

Here's what Scrambled is getting at -- you have to figure out two things: the average number and the number of numbers.  If digit repetition were allowed, the average number would still be the same (because each digit would appear in each place the same percentage of the time), but the number of numbers would increase (since 11111 is now an option and wasn't before).

You can use a similar argument for any question of this type.  For example, can you find 2 different ways to compute the sum of all 4-digit numbers?  (Be careful -- if you want to use this method, it's slightly trickier than similar problems.)

I'm not sure I understood what Scrambled did either.  The best way to do it is this:  the average number will be 1*(average of all possible 1's digits) + 10*(average of all possible 10's digits) + ... to get the overall average.  Make sense?

[quote="Rep123max"]What was confusing me was at first Scrambled had 256 but its 625.  What I dont understand is why the average of 11114, 33334, 44444, 55554, 99994 = the average of all numbers. If it was 4-digit even and the number could be repeated woulnt there be 5*5*5=125 numbers? Andif the digits were say 1, 2, 3, 4, 5, 9 and repetition was allowed and we wanted 4-digits even numbers wouldnt there be a total of 6*6*6*2=432?[/quote]

uh, i was doing it for 5.

But why did you leave the units digit always 4?


Block $B$, with mass $m_{B}$, rest on block $A$, with mass $m_{A}$, which inturn is on a horizontal tabletop. The coefficient of kinetic friction between block $A$ and block $B$ is $\mu_{k}$, and the coefficient of static friction between block $A$ and block $B$ is $\mu_{s}$. A light string attached to block $A$ passes over a frictionless, massless pulley and block $C$ is suspended from the other end of the string. What is the largest mass $m_{C}$ that block $C$ can have so that blocks $A$ and $B$ still slide together when the system is released from rest?

Ok, well the only horizontal force acting on block $B$ is the friction between it and black $A$.  Thus, this is the only accelerating force.  So then we have $m_{B}a_{B}\leq \mu_{s}* N_{B}= \mu_{s}m_{B}g$.

Also, if both blocks are accelerating together, $a_{A}= a_{B}= a_{C}= a$.  The only accelerating force on the 3 block system is the gravitational force on block $C$.  Thus, $m_{C}g = (m_{A}+m_{B}+m_{C})a$.  Therefore, $a = \frac{m_{C}g}{m_{A}+m_{B}+m_{C}}$.

This gives $\frac{m_{B}m_{C}g}{m_{A}+m_{B}+m_{C}}\leq \mu_{s}m_{B}g$.

Now, $\frac{m_{C}}{m_{A}+m_{B}+m_{C}}\leq \mu_{s}$.
Therefore, $m_{C}\leq  \mu_{s}(m_{A}+m_{B}+m_{C})$.
So, $m_{C}(1-\mu_{s}) \leq  \mu_{s}(m_{A}+m_{B})$.

Thus, we finally have $m_{C}\leq  \frac{\mu_{s}(m_{A}+m_{B})}{1-\mu_{s}}$.

What about it mass there is coefficient of kinetic friction $\mu_{z}$ between $m_{A}$ and the table?


geometry
geometric transformation
reflection
symmetry
geometry proposed

Dear Mathlinkers,
let ABC be a triangle , (1) the incircle of ABC, I the center of (1), DEF the contact triangle of ABC,
D' the symmetric of D wrt AI, E' the symmetric of E wrt BI, and E" the symmetric of E' wrt EF.
Prove: D', E and E" are collinear.
Sincerely
Jean-Louis

$ AI, BI, CI$ cut $ BC, CA, AB$ at $ X, Y, Z.$ $ XD', YE'$ cut $ AB$ at $ P, Q,$ respectively. $ APXC$ is kite with $ AP \equal{} AC$ $ \Longrightarrow$ $ \angle QPX \equal{} \angle APX \equal{} \angle XCA \equal{} \angle BCA.$ $ BCYQ$ is kite with $ BQ \equal{} BC$ $ \Longrightarrow$ $ \angle BCA \equal{} \angle BCY \equal{} \angle YQB \equal{} \angle YQP.$ Then $ \angle QPX \equal{} \angle YQP.$ Let $ S \equiv XD'P \cap YE'Q.$ It follows that $ \triangle SQP$ is isosceles with $ SP \equal{} SQ$ and $ (I)$ its incircle tangent to $ PS, SQ, QP$ at $ D', E', F,$ which means $ D'E' \parallel PQ \equiv AB.$ Moreover, its intouch $ \triangle FD'E'$ is also isosceles with $ FD' \equal{} FE'.$ Therefore, $ EF$ bisects $ \angle D'EE'$ internally, which means that the reflection of $ E''$ of $ E'$ in $ EF$ is on $ D'E.$

From my post in this following topic: http://www.mathlinks.ro/viewtopic.php?t=283779, we have known that $ D'E'\parallel{}AB$. But $ AB\perp IF$ $ \Longrightarrow$ $ IF\perp D'E'$, which implies that $ D',E'$ are reflected each other through $ IF$. Therefore $ FD' \equal{} FE'$. But we also have $ E''$ is the image of $ E'$ by the reflection through $ EF$ $ \Longrightarrow FE' \equal{} FE''$ $ \Longrightarrow$ $ FD' \equal{} FE''$. As the result, $ \triangle FD'E''$ is $ F$- isosceles triangle.
$ \Longrightarrow$ $ \angle FD'E'' \equal{} 90^{\circ} \minus{} \frac {1}{2}\angle {D'FE''}$ $ (1)$
But $ \angle D'FE'' \equal{} \angle D'FE \plus{} \angle EFE'' \equal{} \angle DEF \plus{} \angle EFE' \equal{} 90^{\circ} \minus{} \frac {B}{2} \plus{} \angle EFE'$ $ (2)$
Since $ \angle D'FE \equal{} \angle DEF$, $ \angle FD'E \equal{} \angle FE'E \equal{} \angle DEE'$
$ \Longrightarrow 180^{\circ} \minus{} \angle D'FE' \minus{} \angle FD'E \equal{} \angle D'EF \equal{} 180^{\circ} \minus{} (\angle FED \plus{} \angle DEE')$ $ \equal{}$ $ 180^{\circ} \minus{} \angle FEE'$
Hence $ \angle AEF \plus{} \angle E'EC \equal{} \angle D'EF$ $ \Longrightarrow 90^{\circ} \minus{} \frac {A}{2} \plus{} \angle EFE' \equal{} \angle D'EF$
But $ \angle D'EF \equal{} \angle DFE \equal{} 90^{\circ} \minus{} \frac {C}{2}$ $ \Longrightarrow \angle EFE' \equal{} \frac {A}{2} \minus{} \frac {C}{2}$

Thus $ (2)$ $ \Longrightarrow$ $ \angle D'FE'' \equal{} 90^{\circ} \minus{} \frac {B}{2} \minus{} \frac {C}{2} \plus{} \frac {A}{2} \equal{} A$.
Therefore $ (1)$ $ \Longrightarrow$ $ \angle FD'E' \equal{} 90^{\circ} \minus{} \frac {1}{2}\angle D'FE'' \equal{} 90^{\circ} \minus{} \frac {A}{2}$.
But $ \angle FD'E \equal{} \angle FDE \equal{} 90^{\circ} \minus{} \frac {A}{2}$ $ \Longrightarrow \angle FD'E \equal{} \angle FD'E''$, which implies $ D',E,E''$ are collinear. $ \square$

We can assume $ \angle C \equal{} Min\{\angle A, \angle B,\angle C\}$ (other cases are similar)
Denote $ K$ the intersection of $ BI$ and $ EF$. A well-known result: $ \angle BKF \equal{} \frac {\angle C}{2}$
$ \angle E''EE' \equal{} 2\angle KEE' \equal{} 180^o \minus{} \angle C (1)$
$ \angle D'EE' \equal{} \frac {1}{2}\angle E'ID'$
On the other side, $ \angle LID \equal{} 180^o \minus{} \angle FIA \minus{} \angle FID \equal{} \angle B \minus{} 90^o \plus{} \frac {\angle A}{2}$
$ \Rightarrow \angle DID' \equal{} 2\angle B \minus{} 180^o \plus{} \angle A \equal{} \angle B \minus{} \angle C$
Similarly, $ \angle EIE' \equal{} \angle A \minus{} \angle C$
So $ \angle D''EE' \equal{} 1/2(\angle DIE \minus{} \angle DID' \minus{} \angle EIE') \equal{} \angle C (2)$
From $ (1)(2)$ we are done!

Denote the projections $ X$ , $ Y$ of $ C$ on the lines $ BI$ , $ AI$ respectively.
From an well-known property obtain $ X\in EF$ and $ Y\in DF$ .
With symmetry prove easily that $ D'\in YE$ and $ E'\in XD$ . Since
$ \widehat {E''EX}\equiv\widehat {E'EX}\equiv\widehat {YFE}\equiv\widehat {YEF}$ obtain $ \widehat {E''EX}\equiv\widehat {YEF}$ , i.e. $ E''\in ED'$ .


linear algebra
matrix
percent
email
search
induction
probability

Ok, I need help cracking a 2-player, zero-sum, finite, mixed strategy, game theory problem. It's really just a complicated rock-paper-scissors type game, but with a few added conditions.

The object of the game is to have the highest score after 15 rounds. Ties are possible.

Here is the matrix:

[img]http://i545.photobucket.com/albums/hh378/Shalbaal/Matrix.jpg[/img]

First, you need to determine the strength of your opponent. If your opponent is a Beginner then both players can only use Choices [b]A[/b] through [b]H[/b], otherwise you use the entire matrix.

Another condition is that neither player can pick the same Choice twice in a row.

Both players pick simultaneously. Based on the opponent's previous pick I'd like to know what the best responding Choices would be for my next move.

I think the best way to do something like this, and the most ideal, would be to have a simple program where you plug in your opponent's move and it spits out the top 3-5 best responses maybe with percents?

Honestly though I just don't know where to begin on the math side and was hoping someone here might be able to help me out.

Here's the code in list format to copy/paste, I couldn't get it looking good on the screen in the above post, if someone would like I could also email them the excel spreadsheet I originally did this on if the send me a PM with their email.
[hide][list]A	B	C	D	E	F	G	H	I	J	K	L	M	N	O
A	1,1	1,0	0.5,0	0,1	1,0	0,0.5	0,0.5	0,0.5	1,0	0,1	1,0	0,1	0,1	1,1	0,1
B	0,1	0,0	1,0	1,0	0,0.5	1,0	0,0.5	0,0.5	1,0	0,1	1,0	1,1	0.5,0	0,1	1,0
C	0,0.5	0,1	1,1	1,0	0,0.5	0,0.5	1,0	1,0	0,1	1,0	0,0.5	0.5,0	0,1	0.5,0	0,1
D	1,0	0,1	0,1	1,1	0,0.5	0,0.5	0,0.5	1,0	0.5,0	1,0	0,0.5	1,0	0.5,0	0,1	0,1
E	0,1	0.5,0	0.5,0	0.5,0	0.5,0.5	0,0	0.5,0	0,1	0,0.5	0,0	0,0	0.5,0	0.5,0	0,1	0.5,0
F	0.5,0	0,1	0.5,0	0.5,0	0,0	0,0	0,0	0.5,0	0,1	0,0	0,0.5	1,0	0,0.5	0.5,0	0.5,0
G	0.5,0	0.5,0	0,1	0.5,0	0,0.5	0,0	0,0	0.5,0	0,1	0,0.5	0,0	0,1	0.5,0	0.5,0	0,1
H	0.5,0	0.5,0	0,1	0,1	1,0	0,0.5	0,0.5	0.5,0.5	1,0	0.5,0	0,0.5	0.5,0	0.5,0	0,1	0,1
I	0,1	0,1	1,0	0,0.5	0.5,0	1,0	1,0	0,1	0,0	1,0	0,0.5	0,0	0,1	1,0	0,1
J	1,0	1,0	0,1	0,1	0,0	0,0	0.5,0	0,0.5	0,1	1,1	0,0	1,0	0,1	0,1	0,1
K	0,1	0,1	0.5,0	0.5,0	0,0	0.5,0	0,0	0.5,0	0.5,0	0,0	0,0	0.5,0	0,0.5	0.5,0.5	0.5,0
L	1,0	1,1	0,0.5	0,1	0,0.5	0,1	1,0	0,0.5	0,0	1,0	0,0.5	0,0	0.5,0	1,1	0,1
M	1,0	0,0.5	1,0	0,0.5	0,0.5	0.5,0	0,0.5	0,0.5	1,0	1,0	0.5,0	0,0.5	0.5,0.5	0.5,0	0,1
N	1,1	1,0	0,0.5	1,0	1,0	0,0.5	0,0.5	1,0	0,1	1,0	0.5,0.5	1,1	0,0.5	1,1	1,1
O	1,0	0,1	1,0	1,0	0.5,0	0.5,0	1,0	1,0	1,0	1,0	0,0.5	1,0	1,0	1,1	0,0[/list]
[/hide]

Any help is greatly appreciated.

[quote="CluelessonGT"]
Another condition is that neither player can pick the same Choice twice in a row.
[/quote]
This condition screws up the usual minimax theory quite a bit. I'll have to think of what can be done here to incorporate it correctly and not end up with an intractable computational problem.
A good approximation to the truth may be just playing the best strategy for each given round under the condition that neither player can repeat the previous move but this may be suboptimal.

OK, here is what you should do.

1) Learn the basic theory of one round 0 sum games with mixed strategies (I guess it can be found online but I haven't time to search right away. Somebody else will help you here).

2) Now consider the 2-round game. We can reduce it to the one round game in the following way. Take any pair of moves. Look at the second round with these moves removed. Find the corresponding equilibrium strategies. Find the payoffs. Add them to the first move payoffs. You'll get a new matrix of a one round game that is equivalent to the 2-round game in the sense that the expected payoffs are the same if both players use their optimal strategies.

3) Now analyze the game from the end converting the two last rounds into one round in the above way 14 times. You'll get a set of 14 tables telling you what to do (make sure you keep the strategy for the current round and all the strategies for the next "combined" round based on the moves done at the current round.

4) Once you have constructed these tables (it needs to be done only once), just find some decent random number generator and enjoy your computer playing the game for you.

Warning: this is the optimal strategy against a smart opponent who uses the optimal strategy too. If you play against a dumb opponent, you may try to benefit even more on his repeated silliness, but that would be psychology, not mathematics.

Maybe something better can be done than backward induction (described by fedja)... Do you know the origins of this big matrix?

This is unlikely: the "no two identical moves in the row" rule calls for backward induction anyway. Perhaps, if we know the origin of the matrix, we can do it with smaller matrices, but that's about it.

At the risk of asking a dumb question, what does each entry mean?  For example, what does an entry of $ 0.5, 0$ mean?

Literally the matrix is what the game uses for scoring. Player 1 chooses an option A-O. At the same time Player 2 chooses their option.

The numbers beside each are the outcome for scoring. For example if Player 1 chooses A and Player 2 chooses J, then Player 1 scores "0" and Player 2 scores "1." If Player 1 next chooses C and Player 2 chooses E, the score would be 0 to 1.5 in favor of Player 2. The player with the highest score after 15 rounds wins.

As far as the game goes "A through O" are actually attacks, blocks, or dodges. If you block or dodge an attack you get 0.5 points, if you hit the opponent with the attack you get 1 point. There are situations where players both hit each other, where players attacks block each other, or where they both disengage/dodge and nothing happens.

That was my guess of what the entries mean, but then how is it a zero-sum game?  Shouldn't the two numbers in each entry add up to zero (or at least a constant)?

I thought that the task is to get more points than the opponent, so, if $ (a,b)$ is the pair in a certain position, then the corresponding pair of moves is actually just worth $ a\minus{}b$ to the first player and $ b\minus{}a$ to the second one. This means that the matrix can be simplified quite a bit. The alternative interpretation when each player's task is to get the maximum number of points regardless of what the opponent's score is results in the trivial cooperative strategy (A,A), so I decided that this was not what was meant.

Oh, I see, thanks.  So with the first interpretation, each matrix entry may as well be a single number (the difference).

Yes. The interesting question is whether what matters at the end of the game is just winning or the point difference. The strategy discussed so far maximizes the expectation of the point difference but if the task is just to win with the highest possible probability, the analysis should be somewhat different. Then the optimal strategy for each round will depend not only on the last moves but also on the score difference after the previous round, which will result in a few hundred tables instead of $ 14$. Fortunately, it is still a child play for modern computers.

P.S. One rather comprehensive set of lectures on Game theory can be found [url=http://www.math.ucla.edu/~tom/Game_Theory/]here[/url]. That directory seems to be something like a collection of chapters for a book to be written, so you'll have to search through a few files with not very informative names to find what you really need, but a quick glance at them gave me quite a good impression. If somebody knows a better free online source to recommend to the OP, please mention it. :)


If you are submitting a partial solution for a few points, do you need to label it as such? Also, do you need to add some sort of closure to it? Obviously you won't be adding QED, but do you need to add something such as, "This is as far as I got" ?

It's nice to add something like that as a courtesy to the grader, but it's not strictly required.

Hmm, in some competitions I have participated in, pretending to know what I'm talking about when proving something when I really don't pays off.  Like on a Georgia proof contest hosted by Kenessaw State, I bs'd a question and got points for it.  I did put "QED" at the end, so I don't know whether you should or should not.  It depends upon how far you can get.  If USAMTS is anything like USAMO, nontrivial conclusions=partial credit.  Seeing as we still have like over 9000 years until the due date, I suggest thinking more about the problems before submitting or quitting.  You might just get the whole answer that way. :)

Heh Its kinda fun trying to BS a few point out of the USAMO. With 30 minutes left and nowhere near a prospect for a solution, I spent a whole 2 pages writing some BS. I thought I might get points for effort :P

Heh, this might be spam, but I was browsing through the blogs and read that one AoPSer BS'd number six of the USAMO about the mathematicians at the conference by saying "hire a bulldozer to destroy the wall between the two rooms, and there's only one (which is a power of 2, though they did say positive exponents) way to arrange the mathematicians."  :| :arrow:   :rotfl:

But anyways, to the real question posted, I would, if I were in your position, not draw the reader's attention to the fact that it's impossible for you to get a 5 on an incomplete solution. :wink:


I think I have the answer for this but I just want to compare:

Suppose you have 10 indentical red balls, 10 identical blue balls, 10 identical white balls and 10 identical black balls.

a.) How many choices of 6 balls can you make?

b.) How many choices can you make if you must have at least 1 ball of each color?

[hide]a)40!/6!34!=3838380[/hide]

[hide]
combinations with repetitions, C(n-k-1,k)=33!/27!6!=1107568
I'm not exactly sure about this answer, so I wouldn't mind seeing if anyone else got the same answer.[/hide]

[hide="2"]Now I am not sure if this is right but I would like someone to correct my thinking if it is not.  What if you just set aside 4 balls, one of each color, at the beginning and chose any combination of 2 from the last 36.

If this reasoning is sound your answer would be 36 nCr 2.  36! / (2!34!) = 630.  Wow thats totally different than iranman, i think i screwed up big.[/hide]

No, I'm pretty sure I messed up.  I tried using combinations with repetitions, but I'm missing something from my answer.  Don't base your answer on mine because I'm almost positive its incorrect.

I am not sure if iranman's first answer is correct because the balls are identical...

I might be way off
[hide]4 choices of color for each ball... 4^6... 4096???[/hide]

[hide]Must have at least one of each color.. so only 2 balls left to choose which makes 4^2 = 16[/hide]
not sure

For number 1, you could do something like this:

[hide]
let d be any ball at all
d1 d2 d3 ... d6 is our balls
we need to split them into four groups, so we could add lines to break them up

d1 | d2 | d3 ... d5 | d6 is one example of a possible grouping

You're adding three lines, so when you pick the places the lines could be in the answer is
9C2 or 36
[/hide]

For number 2:
[hide]
We set aside four of the balls, so we have only 2 balls to choose the color of. There are only 10 possibilities (I think). I just wrote all of them out...
[/hide]

I'm not sure on either of these either, though...

I think you're right about #2 because the answer would be the 4th triangular number...


Question:
Given a regular polygon of $ n$ sides, which is the probability that, choosing three vertices with random selection, we get a right-angle triangle?

When $ n$ is odd, there aren't any.  When $ n$ is even, we need two diametrically opposed points plus one other.  The probabilities here are not very hard to calculate ;)


AMC AIME geometry ratio inequalities trigonometry algebra

Hi guys,

Since the new school year is coming, this also means new math competitions. To develop our problem skills, I'm going to post some challenging problems and if you get the problem right (by posting in this thread) before other people, you get the point. You must have solutions and the answers in order to get the point.

Note: Only I will post the problems and the new problem will be posted if anyone got the previous problem correct. If no one solved one problem for 3 days, I'll post the solutions and new problems.

This is, as always, open to everybody. Whether your solution is right or not, please post your idea in this thread.

Oh, and none of these problems were made by me. I found these in many sources

[size=150]Problem 1[/size]

What is the smallest positive integer $n$ such that 31 divides $5^n + n$?

[hide="Scores"]

BornForMath: 3
kunny: 2
Altheman: 2
xxreddevilzxx: 2
PenguinIntegral: 1
aidan: 1
paladin8: 1
NoSoupForYou: 1
[/hide]

[quote="Silverfalcon"]What is the smallest positive integer $n$ such that 31 divides $5^n + n$?[/quote]

[hide]well, you want an n so that (5^n+n)=0 mod 31
I will post the cycles of 5^n mod 31 in a minute.

EDIT:25,1,5
So now lets make a table

Powers of 5 mod 31:  5 25 1 5 25 1  5  25  1    5 25    1   5  25   1
n:                                1  2 3  4  5  6  7   8   9  10 11 12 13  14   15
Answer mod 31:          6 26 4 9 30 7 12  2 10  15...............

I need things 1mod3, and n's divisible by 3  are doing that right now. So......
n=30!!!!![/hide]

That's the correct answer.  :)

PenguinIntegral got the 1st point!

[size=150]Problem 2[/size]

Two of the altitudes of the scalene triangle $ABC$ have length 4 and 12. If the length of the third altitude is also an integer, what is the biggest it can be?

Are these AIME level? What level are they?

It looks about AMC 12. This one, at least. :)

I can set up a really messy system of equations.
Trying solving for n in the following:

$144+x^2=(a+b)^2$
$144+y^2=(e+d)^2$
$b^2+16=(e+d)^2$
$a^2+16=(x+y)^2$
$n^2+d^2=(x+y)^2$
$n^2+e^2=(a+b)^2$

I don't know any special properties about the orthocenter that could help me.

[quote="chess64"]Something that could help is that the ratio of the altitude to the side is the same for all sides.

:wallbash_red: I should really stop giving other people hints! :D[/quote]
Thank you!
Then I say the answer is 15.
THANK TOU!!!!!!!!

[quote="Silverfalcon"]15 is not the correct answer.

Note: Please do not give the hints away. If you think you know how to solve the problem, why not just solve it?  :D

These problems are below AIME level. But it will be AMC-12 levels and I'm trying to post the ones that are in level of later ones.[/quote]

Assume the common ratio is one(it could be any number and not change the answer, right?

then one side is 12, on 4, and one x. Maximising x with the triangle inequality we get 15.

No, your argument has problem.

Although triangle inequality states $a < b+c$, this doesn't mean for height (or altitude) $h_a < h_b + h_c$.

[quote="PenguinIntegral"]Assume the common ratio is one(it could be any number and not change the answer, right?

then one side is 12, on 4, and one x. Maximising x with the triangle inequality we get 15.[/quote]

Assume the common ratio is 2(it could be any number and not change the answer, right?

then one side is 24, on 8, and one x. Maximising x with the triangle inequality we get 31.

But I assumed a commin ration one between the sides and altitudes(does it change the answer to have different ratios) :oops:

There is a very well-known inequality for heights using triangle inequality.

See if you guys can find it. This is very useful inequality because as you see, this problem was pretty tough to solve without it (and obviously it should since it was #29 in AHSME).

BornForMath, there are not 4131 positive 4-digit integers starting with 1...

[hide="Solution"]
[b]For only one digit being repeated:[/b]
Let repeated digit is not 1.
Then we can have number of such numbers $=9\cdot3\cdot 8=216$
When repeated digit is 1.
Then number of such numbers $=3!\cdot \binom{9}{2}=216$

[b]When 2 digits are repeated:[/b]
Number of such numbers $=3\cdot\binom{9}{1}=27$

Hence the total number of solution is $216+216+27=459$ [/hide]

[quote="daveed"][hide]
looking at the cases...
Only the last digit of anything counts

7^1 ends in 7
7^2 ends in 9
7^3 ends in 3
7^4 ends in 1
7^5 ends in 7

So 7^7 ends in 9.

find out what 9^7 ends in...
it goes
9^1 - 9
9^2 - 1
...
meaning 9^7 ends in 9
and because everything is raised to the last power, it's got to end in 9.
[/hide][/quote]

daveed,
you got $7^5$ ends in 7
$\Rightarrow 7^6$ ends in 9 and not $7^7$.

Got it ;)

haha, oops  :blush:

[hide="then this might be it..."]

So 7^7 ends in 3
working with 3, there's
3^1 - 3
3^ 2 - 9
3^3 - 7
3^4 - 1
...
3^7 - 7.

So it's going to oscillate between 3 and 7.  If there's an odd number of 7's in the exponent, it's going to return a 3.  If there's an even number, we'll get a 7.
Varun, when you say there are 1001 7's, I'll assume you mean including that 7 which isn't an exponent... meaning that there's 1000 7's as exponents, which is evin...

meaning we get $\boxed{7}$ ?
[/hide]

[quote="chess64"][quote="varun"][b]Problem 19[/b]
What is last digit in the number $((...(((7)^7)^7)...)^7)$?
Total $7$'s are 1001.[/quote]

Huh? :huh: Why did you post the next problem? This isn't a marathon, it's just Silverfalcon posting some problems. Unfortunately, he was unable to post any more problems, so I started doing it. Oh well, I'll post problem 20. ;)[/quote]

I'm really very sorry. I should have read the first page.
Now that I've posted it either you should delete it or you should reply to it.

starting from the well known fact that
7^1=7
7^2=49
7^3=..3
7^4=..1
7^5=..7
we can say that 7^(4k)=..1 ; 7^(4k+1)=..7 ; 7^(4k+2)=..9; 7^(4k+3)=..3
since (a^m)^n=a^(m*n) we get 7^(7*7*..*7).the fact that we have a total of 1001  7s is not very clear to me (wether i should count the one in the basis or not). if we do count it, we'll have 7^(7^1000).since 7^1000 is the perfect square of 7^500, it has the form of 4k+1, and 7raised to such a power will end in 7(mod 10 :P).
on the other hand, if we have 7^(7^1001), 7^1001=(7^500)^2*7=4p+3, and now the final digit will be 3 (mod 10 :P).
please correct any mistakes and sort out my inclarity.

The last digit will be 3 mod 4, from your reasoning. We need it mod 10.

[size=150]Problem 20[/size]

Find $\Im(\cos 12^\circ + i\sin 12^\circ + \cos 48+i\sin 48^\circ)^6$.

[quote="10000th User"]One quick question: what is $\Im$? :?[/quote]

Imaginary part of complex number. :)


Can anyone solve or explain why the following problem cannot be solved?

Given ten numbers, let them be 50, 43, 1, 2, 3, 7, 5, 4, 4, 1, how can we decide with 99% confidence that the difference between the first two numbers is insignificant relative to the comparison with the first two numbers and the others (1, ..., 4)?

I hope the problem is clear.

Define 'insignificant'

Plus, how are these 10 numbers chosen?

Insignificant is supposed to be something like, they [b]could be the same number[/b] because relative to the other numbers their difference is really small.

As for the second question, I don't really know. Does it have to be put in context? What value does sampling information add?

An example of a good problem is:
Five numbers randomly choosen from distribution $A$ are 2,4,5,6,3 and five random numbers from $B$ are 4,5,6,7,3.  With what confidence can we say that $A$ and $B$ have  (or don't have) the same mean? What about variance?


trigonometry
function
real analysis
real analysis unsolved

In the  $L_{[-\pi;\pi]}^{2}$ normed space, let  $A$ be the smallest closed subspace of $L^{2}$ that contains $\frac{1}{2},\cos x,\cos 2x,...,\cos nx,...$ and $B$ be the smallest closed subspace that contains $\frac{1}{2},\sin x, \sin 2x,...,\sin nx,...$.Prove that \[\sum_{n=1}^\infty \frac{\sin (n+\frac{1}{2})x}{n}\notin A+B\]

Can anyone hepl me? I've tried to solve this pro but it seems  too difficult to me. :(

Hmm... for one thing, $A+B$ is all of $L^{2}(-\pi,\pi).$ (It also doesn't seem necessary to include the constant in $B.$ All that does is give $A$ and $B$ a nontrivial intersection.)

So now we have $f(x)=\sum_{n=1}^{\infty}\frac{\sin\left[\left(n+\frac12\right)x\right]}n$

If $f$ is in $L^{2},$ then $f\in A+B.$ So that's the only remaining question: is it square integrable?

The computation is helped by the fact that the terms of that sum are orthogonal. Hence we can calculate that

$\|f\|_{2}^{2}=\pi\sum_{n=1}^{\infty}\frac1{n^{2}}$

Hence $f\in L^{2}=A+B.$

In fact, since $f$ is odd, we have that $f\in B.$ (That is, it can be represented by a Fourier sine series with square summable coefficients.)

Can you show me why $A+B=L^{2}$ and $f\in B$? Thanks.

It is known that the closure of the span of $\{1,\cos x,\sin x,\cos2x\sin2x,\cos3x,\sin3x,\dots\}$ is all of $L^{2}.$ That's a very central theorem in Fourier analysis, and deals with some of the most important basic princibles of Hilbert spaces. Another way of phrasing that: this is a complete orthogonal set, which means that it is not possible to find a nonzero function which is orthogonal to every member of this set. For the proof, see any good development of Fourier series.

Now, take an arbitrary $L^{2}$ function. It can be represented as a Fourier series which converges in the $L^{2}$ sense. Take that series and separate into two series: a cosine series and a sine series. Each of these series converges. The cosine series converges to a member of $A$ and the sine series converges to a member of $B.$ Hence $f\in A+B.$

Another way to get the decomposition into $A+B$ is simply to decompose a function into its even part and odd part. This particular $f$ belongs to $B$ simply because it is odd.


geometry
geometric transformation
reflection
circumcircle
perpendicular bisector
geometry unsolved

It is clear that the magnitude of the angle $ U_{a}AT_{a}$ is $ \pi/2$, and thus, we obtain the vertex $ A$ by intersecting the perpendicular through $ H$ to $ U_{a}T_{a}$ (which is the line $ BC$) with the circle with diameter $ U_{a}T_{a}$. On the other hand, the reflection $ H'$ of $ H$ into $ U_{a}T_{a}$ lies on the circumcircle of $ ABC$, $ AH$ being in this case a chord in the circumcircle $ \Gamma$ of $ ABC$. But, $ H$ and $ O$ are isogonal conjugates, and so $ O$ lies on the reflection of $ AH$ into $ AT_{a}$. Thus, we obtain the circumcenter $ O$ of $ ABC$ by intersecting the reflection of $ AH$ into $ AT_{a}$ with the perpendicular bisector of $ AH'$. Therefore, we can now construct $ B$ and $ C$ by intersecting $ \Gamma$ (which is now constructible since we have its center and $ A$) with the line $ U_{a}T_{a}$.


function
logarithms
complex analysis
complex analysis unsolved

If f is analytic at the origin and f ' (0) does not equal to 0, prove the existence of an analytic function g such that f(z^n) =f(0)+(g(z))^n in a neighborhood of 0.

Thx in advance

Write $f(x)=f(0)+x h(x)$ for some function $h$. Then $h(0) \neq 0$, thus you can take (some) logarithm of $h(x^n)$ in some neighbourhood of $0$. Now set $g(x) = x e^{\frac{1}{n} \ln ( h(x^n) )}$.


For every positive integer $ n$, prove the inequality:

$ \frac{3}{1!\plus{}2!\plus{}3!}\plus{}\frac{4}{2!\plus{}3!\plus{}4!}\plus{}...\plus{}\frac{n\plus{}2}{n!\plus{}(n\plus{}1)!\plus{}(n\plus{}2)!}<\frac{1}{2}.$


function
calculus
integration
trigonometry
algebra unsolved
algebra

Well a very interesting & adorable fact: :)

[color=blue]problem[/color]:
Let $f: \mathbb{N} \longrightarrow \mathbb{R}$ and

$f(n)=\sum_{k=1}^n (\frac{|Sin(k)|}{k})$.

1.Let $k \in \mathbb{R}$,Is $f$ bounded?(This one is self-designed :) )
2.Let $k \in \mathbb{N}$,is $f$ bounded?

2) Seems convergent by Abel's rule, so in particular it is bounded.
What is 1) supposed to mean ? Integration ?

Well,i just saw the second problem somewhere and Designed the first one (i have said its self designed  :) )
I havent studied integration yet so i cant say anything but do we need it here? ,i just asked is it bound less(the first problem,We can just sum them up ,cant we ? do we need integration)) and i think ,yes its is bound less :)

for two :Really its bounded ?Can you prove  it.

You were wrong its not bounded .
I have managed to prove the second one :

[color=blue]Solution[/color]
From every 3 natural number lik $n,n+1,n+2$ one of the $|Sin(n),Sin(n+1),Sin(n+2)|$ is bigger than $\frac{1}{2}$
So :

$\sum_{n=1}^{k}| \frac{Sin(n)}{n}| \ge \frac{1}{2} \sum_{i=1}^{[\frac{k}{3}]} \frac{1}{3i}$
but we know that right side is boundless so the left is bound less two.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=62904

And the first question makes no sense.  If you have a sum indexed by natural numbers, you cannot later set them to real numbers.

If you are asking whether

$\int^{\infty}_0 \big|\frac{\sin x}{x}\big|dx<\infty$, the answer is no, since you can compare it to $\int^{\infty}_0 \frac{dx}{x}$

[quote="blahblahblah"]
And the first question makes no sense.  If you have a sum indexed by natural numbers, you cannot later set them to real numbers.
[/quote]

Well i think you are right ,its non_sense (It was self_designed ,i just though why shouldnt we change the indexe to real numbers  :oops: terribly sorry)

Sorry I had read $sin(n)$, in which case Abel's rule applies. In the case of $|sin(n)|$, it doesn't apply anymore.


Why does the forum change colors once in a while?

I have yet to experience that. Browser version, OS version please.

This happened on both my XP a couple months ago and just now with my Vista.  The browser I have is IE7.

EDIT To clarify this happens to the entire page when it loads. After navigating the page or refreshing a few times, the color goes back.

It happened again. Probably something to do with the browser.

Very likely it is a browser issue, and not something that I can help with.


Ring Theory
number theory solved
number theory

Let d be a positive integer,and S be the set of all positive integers of  form $x^2+dy^2$ where x and y are non-negative integers.

Prove that if $a,p\in S$,such that p is a prime and p|a,then $\frac{a}{p}\in S$

[quote="ZetaX"]I think it's wrong (even if you want some extra condition):
Take $d=5$, then $21$ can be represented, but $7$ not.[/quote]
but $3$ cannot be represented...

[quote="wei_wenzhe9170"]just use
(x^2+dy^2)(z^2+dw^2)=(xz-dyw)^2 + d(xw+yz)^2[/quote]

And your solution are...?Please!

Sorry for my nonsense above  :blush:

Let $p=x^2+dy^2$ and $a=z^2+dw^2$.
Note that $p \nmid x,y$ since that would give a contradiction.
We have also $p=x^2+d(-y)^2$. By using numbers of the form $x+\sqrt d y$ we get the formula already mentioned.
This leads to $pa=(xz-dyw)^2+d(xw+yz)^2=(xz+dyw)^2+d(xw-yz)^2$.
Now $p|wz(x^2+dy^2)+xy(z^2+dw^2)=(xz+dyw)(xw+yz)$, so $p|(xz+dyw)$ or $p|(xw+yz)$. Similar, $p|(xz-dyw)$ or $p|(xw-yz)$

1. case: $p|(xw+yz)$
this gives that $p|(xz-dyw)$ and dividing we will get that $\frac{a}{p} = \left( \frac{xz-dyw}{p} \right)^2 + d \left( \frac{xw+yz}{p} \right)^2$.
2. case: $p|(xw-yz)$
this gives that $p|(xz+dyw)$ and dividing we will get that $\frac{a}{p} = \left( \frac{xz+dyw}{p} \right)^2 + d \left( \frac{xw-yz}{p} \right)^2$.
3. case: $p | (xz+dyw)$ and $p|(xz-dyw)$.
this gives that $p|2xz$, but $p \nmid x$, so $p | 2z$. If $p|z$, then $p|w$ too, and we get that $\frac{a}{p} = \left( \frac{xz-dyw}{p} \right)^2 + d \left( \frac{xw+yz}{p} \right)^2$ works. So we are left with $p=2$, but that's just possible for $d=1;2$, for which we show it directly.

In more general manner, it can (up to some calculation) easily be proved using prime ideal decomposition and norms (that's also where this nasty case $p=2$ comes from).

[quote="nguoikhonkho"]can use complex number[/quote]
This post is complete useless: either give your solution, or don't post this.

But the only purely non-calculating solution I know would use non-elementary algebraic number theory... so please post yours...


"The Surface area, including top and bottom, or a rigth circular cylinder is 950 sq ft. Express the volume of the cylinder as a function of r."

In this problem you're only suppose to have R's and numbers in the volume of a cylinder right? like the height and pi must be in form of R's?

$\pi$ can still be $\pi$, but it just means you can't have any other unknowns in it other than $r$.  No heights.

It looks right to me.  Just simplify more.

[quote="Farcus"]I come up with

pi * r sqaured *((950-2pi rsquared)/2pi r) but it says I am wrong? how could this be I figured out H .[/quote]

so this would be:

$\pi r^{2}(\frac{950-2\pi r^{2}}{2\pi r})$
$r (\frac{950-2\pi r^{2}}{2})$
$r (425-\pi r^{2})$

[quote="Farcus"]
In this problem you're only suppose to have R's and numbers in the volume of a cylinder right? like the height and pi must be in form of R's?[/quote]
Pi is not a variable, and if you can express it in terms of r, wow, good for you, but you don't have to. pi is like 12, or 23... it's a constant.