Somehow, this reminds me of the Monty Hall problem.

It is exactly the Monty Hall problem. If you've never heard of it before please don't google before you try to answer the question because it's a really fun one to answer yourself  :P.

[quote="Cartman"]Suppouse you are on a gameshow and can choose one of three doors. Behind one door is a prize. The rest is empty. You choose a door. Then the host, who knows where the prize is show you an empty door (something he would have done anyway). Then there are two doors left. The host then offer you to change to the other door. Will that be to your benefit? Is there a greater chance that the prize is behind the other door?[/quote]
It would be better to switch.
[hide="Reason Why"]
You should choose the other door, because if the one you first chose is right, you would have a 1/3 chance, because you would have to be right from the beginning. That leaves the other door with a probability of 2/3.[/hide]

For more detail, visit
[url]http://www.cut-the-knot.org/hall.shtml[/url].
It's a really great website.

[hide="Another approach"]Suppose Monty Hall wasn't there to open one of the empty doors.  Then do some casework and think about what switching would do.[/hide]

[hide]The probability that your first guess is correct is $\frac13$.  The probability that you were not correct on your first guess is $\frac23$, making the probability of you winning by switching $\frac23$.[/hide]


algebra
polynomial
modular arithmetic
number theory unsolved
number theory

Find all natural numbers $ n$ for which $ 2^n\minus{}n^2$ is divisible by $ 7$.

This is much easier than the Ramanujan-Nagell equation.  Exponentials and polynomials share the property of being periodic modulo a prime.

Is it just me, or is the Ramanujan-Nagell equation and this problem very different problems.

[hide]
We are looking for $ n \in \mathbb{N}$ such that $ 2^n\minus{}n^2 \equiv 0\pmod{7}$.

Since $ 2^3 \equiv 1 \pmod{7}$, the residue of $ 2^n$ in $ \pmod{7}$ is periodic with period $ 3$.

Since $ 7$ is prime, the residue of $ n^2$ in $ \pmod{7}$ is periodic with period $ 7$.

Thus, the residue of $ 2^n \minus{} n^2$ is periodic with period $ lcm(3,7) \equal{} 21$.

Note that $ 2^n\minus{}n^2 \equiv 0\pmod{7} \Longleftrightarrow 2^n \equiv n^2 \pmod{7}$

List out the residues in $ \pmod{7}$ of $ 2^n$ and $ n^2$ for $ n \in \overline{0,20}$:

$ 2^n \equiv 1,2,4,1,2,4,1,2,4,1,2,4,1,2,4,1,2,4,1,2,4 \pmod{7}$

$ n^2 \equiv 0,1,4,2,2,4,1,0,1,4,2,2,4,1,0,1,4,2,2,4,1 \pmod{7}$

So, the natural numbers $ n$ such that $ 2^n \equiv n^2 \pmod{7}$ are:
$ n \equiv 2,4,5,6,10,15 \pmod{21}$[/hide]

I don't know, this site tell me:

http://mathworld.wolfram.com/RamanujansSquareEquation.html

That is [b]not the equation we're looking at.[/b]

$ 2^n$ modulo 7 has period of 3. $ n^2$ modulo 7 has period of 7. So $ 2^n\minus{}n^2$ modulo 7 has period of 21. It is easy to check $ n\equal{}1$ to 21. :D


This is most likely written somewhere blindingly obvious, but I can't find it.

When will round two problems be released?  Exactly one month (10-22) before the due date (11-22)?

Offtopic:
I'm ready to get started.  It's been lots of fun, and I'm starting to learn LaTeX (with VIM, Konsole, and Ratpoison on Linux), so the next set can be typeset properly instead of written with OpenOffice.org Math Writer.  After the tenth or so imbedded object, OO.o tends to start slowing down, which isn't fun.

Edit:  Thanks.  I have joined the mailing list, but I like advanced warning.

Glad you're enjoying it.  Round 2 will probably be up a few days before that.  Join the Mailing List (click Stay Informed on the menu at left) to be informed of when they are released.


I   want   that   there   are    no   page   number  on   the   first   page  of  each   part   , but   page   number    on    the   first   page  of  each  chaper(actually,  footer  of   the   first   page  of  each  chaper     contain  page   number  centered)

How  to  get   that?

Thanks  a  lot!

[code]\documentclass[openany]{book}
........[/code]

Redefine \part as you did before to include \thispagestyle{empty}

It  seems  that  the  following  codes  are no differences:

[code]\let\oldpart\part
\let\oldchapter\chapter

\def\part
  {\cleardoublepage
   \def\chapter
     {\cleardoublepage
      \let\chapter\oldchapter
      \chapter}%
   \oldpart}[/code]

[b][color=red]and[/color][/b]

\let\oldpart\part
\let\oldchapter\chapter

\def\part
  {[color=red]\thispagestyle{empty}[/color]
   \cleardoublepage
   \def\chapter
     {\cleardoublepage
      \let\chapter\oldchapter
      \chapter}%
   \oldpart}

[code].........
\part{Title of this part\thispagestyle{empty}}
.......[/code]

run  pdflatex  once,  get    some  warnings!

run  pdflatex  twice,  get  hundreds  mistakes!

It must be something else causing the problem
[code]\documentclass[openany]{book}
\begin{document}
\part{One\thispagestyle{empty}}
\end{document}[/code]doesn't have any errors.

[b][color=red]Redefine \part [/color][/b] is  the  best way, in my opinion

But  I  don't  know   how   to  write  the  code................. :(

The problem is that you were not showing all your code - you were using \frontmatter and \mainmatter which cause the conflict. So use
[code]{\makeatletter
 \let\ps@plain\ps@empty [/code]instead.
In future you must give a [b]complete, short[/b], [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=157945]minimal example[/url].

How  about  the  following  code?
 ugly  code !
[code]
\clearpage
\makeatletter
\let \asas \ps@plain
\let \ps@plain \ps@empty
\makeatother

\part{Test}

\makeatletter
\let \ps@plain \asas
\let\asas\relax
\makeatother

\chapter{}
.......
\clearpage
\makeatletter
\let \asas \ps@plain
\let \ps@plain \ps@empty
\makeatother

\part{Test}

\makeatletter
\let \ps@plain \asas
\let\asas\relax
\makeatother

\chapter{}
.......
[/code]

[code]\makeatletter
\let\s@endpart\@endpart
\def\@endpart{\thispagestyle{empty}\s@endpart}
\makeatother
[/code]


trigonometry
geometry
geometric transformation
rotation
circumcircle

If we have ABC echilateral, with A(a), B(b), C(c) is this ecuation true:
 $ a \equal{} \epsilon b$, where $ \epsilon \equal{} cos\frac {2\pi}{3} \plus{} i sin\frac {2\pi}{3}$?
If yes, then why?

a = /epsilon * b
1 = /epsilon

/epsilon = cos (2pi/3) + isin (2pi/3)
          1 = -1/2 + i (sqrt(3)/2)

They aren't equal so it doesnt work. unless im mistaken with what that little sign that looks like a c.

Let triangle $ ABC$ be on the complex plane with C being the origin.

 [geogebra]1db05db85e9f96b999b48c683f36aa1bd36853a1[/geogebra]

It is known that if point C is the rotation of B with respect to A by the angle $ \theta$, we have $ c\equal{}a\plus{}(b\minus{}a)\text{cis}\theta$.

In this case, point A is the rotation of B with respect to C by $ \frac{\pi}{3}$.

Thus, we have $ a\equal{}c\plus{}(b\minus{}c)\text{cis}\frac{\pi}{3}$

However, since $ C$ is origin, we have $ c\equal{}0$.

Thusly, $ a\equal{}b\text{cis}\frac{\pi}{3}\equal{}b(\cos\frac{\pi}{3}\plus{}i\sin\frac{\pi}{3})$.

Hence, we conclude that the expression is not true.

[quote="FantasyLover"]Let triangle $ ABC$ be on the complex plane with C being the origin.[/quote]
Nope; what you want is for their circumcenter to be the origin.  Then multiplication by $ \epsilon$ is equivalent to rotation by $ 120^{\circ}$.


modular arithmetic
algorithm
function
number theory
totient function

Determine the last digit of $ 3{}^3{}^3$.

how do you do this?

Let's see if we can determine a pattern in units digits.

$ 3^1 \equal{} 3$

$ 3^2 \equal{} 9$

$ 3^3 \equal{} 7$

$ 3^4 \equal{} 1$

$ 3^5 \equal{} 3$

$ 3^6 \equal{} 9$

$ 3^7 \equal{} 7$

...

Aha! There is a pattern! The units digits go like (3,9,7,1) and then repeat. Since $ 33 \equiv 1 \pmod 4$, $ 3^{33}$ has a units digit of $ \boxed{3}$.

what about $ 3^{3^3}$

please explain without the use of modulo numbers.

You use the same method as isabella did. The units digits of the powers of three cycle every 4 powers and the division algorithm gives us

$ 3^3 \equal{} 27 \equal{} 4*6 \plus{} 3$

So that means that the last digit of $ 3^{3^3} \equal{} 3^{27}$ will be the last digit of $ 3^3$ which is $ \boxed{7}$.

The point in finding the pattern is seeing that it cycles through 4 times before going back to the one it started in. In here the pattern is that the last digit is 3 when 3 is elevated to a power in the form of $ 4k \plus{} 1$, were $ k\ge0$. This is were modulus comes in, $ 4k \plus{} 1\equiv 1\pmod{4}$ therefore if 3 is elevated to a power that is congruent to 1 $ \pmod{4}$ then it ends in a 3.

So to find out the last digit of $ 3^{3^3}$ we know that $ 3^3 \equal{} 27$, so $ 3^{3^3} \equal{} 3^{27}$

And from the previous pattern we see that
$ 3^{4k}$ ends in a $ 1$
$ 3^{4k \plus{} 1}$ ends in a $ 3$
$ 3^{4k \plus{} 2}$ ends in a $ 9$
$ 3^{4k \plus{} 3}$ ends in a $ 7$

Since $ 27 \equal{} 4\cdot 6 \plus{} 3$, $ 3^{27}$ it ends in a $ 7$

Edit: Beaten, by alot

Or you can be cheap and kill it with [url=http://mathworld.wolfram.com/TotientFunction.html]Euler's totient function[/url].

$ 3^{\varphi(10)}\equal{}3^4\equiv 1\pmod {10}$, and $ 33\equal{}32\plus{}1\equiv 1\pmod 4$, so our answer is $ 3^1\equal{}\boxed{3}$.

$ 3^3\equiv (\minus{}1)^3\equal{}\minus{}1\equiv 3\pmod 4$, so the answer to the other problem is $ 3^3\equal{}27\equiv \boxed{7}\pmod{10}$.


induction
algorithm
probability
analytic geometry
geometry
invariant
superior algebra

Don't you find frustrating that you can find or read a solution and have the feeling that a better solution can be done?
For example, the following problem:
If we have the numbers 1, 1/2, 1/3, ..., 1/100, and pair at a time, we erase a and b and replace them with the single value a+b+ab, what will the final result be and why?
For which I gave this solution:
[hide]we can prove by induction
P(n): given any n numbers a_1,a_2,...a_n, replace each pair x,y with x+y+xy, we obtain the final result (a_1+1)(a_2+1)...(a_n+1)-1

step 1 is obvious: the final result is a_1+1-1=a_1

step P(n)--> P(n+1)
Suppose that P(n) is true
suppose we have the list of numbers a_1,a_2,..,a_n,a_{n+1} and we want to prove P(n+1) for these numbers.
Suppose that we applied the algorithm for n numbers, we obtain the intermediary result m and we want to pair m with the last number from list. Without loss of generalization, we can assume that the last number from list a_1,..,a_{n+1}
to be paired is a_{n+1}
Since m is the final number after the algorithm was applied to the first n numbers, by induction hypothesis, we have:
m=(a_1+1)(a_2+1)...(a_n+1)-1

Therefore, the final result after the algorithm is applied to n+1 numbers is:
m+a_{n+1}+ma_{n+1}= (a_1+1)(a_2+1)...(a_n+1)-1+a_{n+1}+
+[(a_1+1)(a_2+1)...(a_n+1)-1]a_{n+1}=
=(a_1+1)(a_2+1)...(a_n+1)+(a_1+1)(a_2+1)...(a_n+1)a_{n+1}-1
=(a_1+1)(a_2+1)...(a_n+1)(a_{n+1}+1)-1
We proved P(n)-->P(n+1)

P(1) true, P(n)-->P(n+1) true
Then the induction worked, so P(n) is true

In our case we have a_1=1,a_2=1/2,...,a_n=1/n
we obtain the final result:
(1+1)(1+1/2)...(1+1/n)-1=
=2*3/2*4/3*...*(n+1)/n-1
after cross multiplying terms, we obtain the final result
= n+1-1=n

ANSWER:n
[/hide]

The solution is there but it doesn't really explain why it works. I have the feeling that many induction proofs are "cheating". They work out logically but don't explain what is behind the scenes.

Another example, this one from probability. The question was asked today on YA forum:
Two numbers, x and y, are chosen from the interval (0,3). What is the probability that a triangle with sides of 1, x, and y exists?

The answer, given by user "Yes":

[hide]A triangle can be made from 3 lengths, if and only if the sum of any 2 of them is longer than the third. So for x,y,1 we should have:

x+y>1
x+1>y
y+1>x

Consider an XY plane, and consider that the (x,y) we choose is a point on the plane. It will be in a square of length 3 that goes from 0 to 3 in the x coordinates and from 0 to 3 in the y coordinates.

Then just look at the regions in the plane that satisfy each of these inequalities. (i.e. draw the line that you get by changing the > to = in the equations above, and see what side of the line is valid, shade out the invalid parts.)

Then it's just a matter of dividing the valid area by the total area (9). You should get 4.5, so the final probability should be 0.5.[/hide]

The result is 1/2. Isn't it too beautiful to be true? Isn't it other, more advanced reason, why the result is exactly 1/2?

I absolutely agree; that's why I think it's always good to try to find multiple solutions to a problem to better get at what's really going on.

For example, your first proof is phrased in terms of induction but it's really an invariance proof:  the quantity $ (a_1 \plus{} 1)(a_2 \plus{} 1)...(a_n \plus{} 1)$ is invariant and can be computed for the initial case, therefore known for the final case.  The erasing step is set up so that this is true.

As for the second problem:  probabilities of $ \frac {1}{2}$ generically suggest a bijective proof.  That is, it should be possible in theory to find a bijection between the cases that work and the cases that don't work.  I haven't plotted out the area in question; does it have any nice symmetries that suggest that this would be the case?  (It's possible the answer is a coincidence.)

For the first one, the reason behind this is a formal group law $ f(x,y) \equal{} xy \plus{} x \plus{} y$. See http://en.wikipedia.org/wiki/Formal_group .

For the second one, probability is by definition (!) finding some measures; so this problem is already a problem of measure (=area or volume) by defintion. See http://en.wikipedia.org/wiki/Measure_(mathematics) . Probability has two major measures: Counting and Volume; the first one is the known "law" that if we choose some $ a$ from a finite set $ M$, then the probability of $ a\in A$ is exactly $ \frac {|A|}{|M|}$, the second one is what is used in the problem.
There cannot be some direct bijective proof, as both sets are infinite; one needs to look at measure-preserving maps.


The square loop  is made of wires with total series resistance 10.0 Ohm
. It is
placed in a uniform 0.100 T magnetic eld directed perpendicularly into the plane of the
paper. The loop, which is hinged at each corner, is pulled as shown until the separation
between points A and B is 3.00 m. If this process takes 0.100 s, what is the average
current generated in the loop? What is the direction of the current?

Are you posting these problems just to give others something to think about, or are they problems that you want to solve yourself but need help with?  In the latter case, you should post what progress you've made rather than just the problems (per the rules for posting on the physics forum).

[quote]These problems could seem easy for you,but you are not the ony person in this forum.These are problems just for practise.I don't need thier solutions.So if you want to solve any of them,you should solve ,not to reply with such unpleasant posts.[/quote]

I didn't mean to be unpleasant, and I'm not saying that your problems are too easy - I actually think they're quite interesting.  I just wanted to make sure that you weren't posting homework problems or something.


abstract algebra
superior algebra
superior algebra solved

Let $X,Y,S$ be noetherian schemes. Is $X \times_{S}Y$ again a noetherian scheme?

Cover $S$ by a finite number of affine open subsets that must be spectra of noetherian rings. Call these open subsets noetherian affines and denote them $S_{i}$
The preimages of every noetherian affine $S_{i}$ in $X$ and $Y$ are finite unions of noetherian affines $X_{ij}$ and $Y_{ik}$.
By the glueing construction, the problem reduces to:

Are fibre products $X_{ij}\times_{S_{i}}Y_{ik}$ noetherian if all the terms involved are noetherian?
In algebra: If $A,B,C$ are noetherian rings such that there exist ring morphisms $f: A\to B$ and $g: A\to C$ i.e. $B,C$ are $A$-algebras, is $B\otimes_{A}C$ noetherian?

The answer is no, an example is $\mathbb C\otimes_\mathbb Q\mathbb C$.

(More generally, if $K$ has characteristic zero, then setting $I=\ker(L\otimes_{K}L\to L)$ it is well-known that $I/I^{2}=\Omega_{L/K}$ has dimension equal to the transcendence degree of $L/K$. If this is infinite, $I$ cannot be finitely generated.)

[quote="owk"](More generally, if $K$ has characteristic zero, then setting $I=\ker(L\otimes_{K}L\to L)$ it is well-known that $I/I^{2}=\Omega_{L/K}$ has dimension equal to the transcendence degree of $L/K$. If this is infinite, $I$ cannot be finitely generated.)[/quote]

I cannot find the result you state without the additional condition of $L$ being a finitely generated extension of $K$. Where did you find it?

Bourbaki, Algebra, Chap. V, §16, No. 4, Corollary of Thm. 2


calculus
integration
function
analytic geometry
trigonometry
absolute value
real analysis

I have a function in spherical coordinates that's to be integrated over whole space. I just can't get the correct answer, which is $5a/4$. Anyone willing to work out a solution?

$\int_{V}e^{-4(r_{1}-r_{2})/a}\frac{1}{\sqrt{r_{1}^{2}+r_{2}^{2}-2 r_{1}r_{2}\cos(\theta_{2}) }}r_{1}^{2}\sin(\theta_{1}) r_{2}^{2}\sin(\theta_{2}) dV$

where $dV$ is $d\phi_{1}d\theta_{1}dr_{1}d\phi_{2}d\theta_{2}dr_{2}$

The book suggests doing the $\theta_{2}$ intergral first, then the $r_{2}$ integral (since it's an integral covering the whole space, there shouldn't be any problems with changing the order of integrations). Because absolute value of $r_{1}-r_{2}$ occurs ($\sqrt{(r_{1}-r_{2})^{2}}$), it has to be done in 2 parts, one ranging from $0 \to r_{1}$ and other is between $r_{1}\to \infty$ --boundaries of the latter one can be extended to $0 \to r_{1}$ for it can be transformed into gamma function with a simple change of variable, we can then subtract away the $0 \to \r_{1}$ part, which seemed to be easier to compute.

whoops. some terms were missing, i've corrected the integral.


limit
advanced fields
advanced fields unsolved

$ \lim_{x\rightarrow\infty} x\bigg[\bigg(1\plus{}\frac{1}{x}\bigg)^x\minus{}e\bigg] \equal{} ?$
show steps

hello, the limit is $ \minus{}\frac{e}{2}$.
Sonnhard.


geometry
blogs
AMC
AIME
probability
trigonometry
3D geometry

Rules:

1. Keep it MOEMS level.
2. When you answer, give a new problem. You can always make up one, or use Google.

Problem One:

Sally is thinking of a number. The product of the digits is one half of the number. What is the smallest positive, two-digit integer she could be thinking of?

call the tens digit a and the ones digit b. then the numerical value is 10a+b

now we know that $ ab \equal{} \frac{1}{2}(10a \plus{} b)$ so $ 2ab \equal{} 10a \plus{} b$ now we knwo b is even

the smallest number satisfying that is 36

new prob

find the hypontenuse of a triangle with legs 20 and 21

We use the Pythagorean Theorem.

$ 20^2 \plus{} 21^2 \equal{} 841$

$ \boxed{29}$

New Problem:

Similar to the first problem --

Randy wants to find the smallest, positive, two-digit integer which is twice the sum of its digits. What is it?

wth?? ernie's marathon got locked too??

let the 2 digit number be xy.

$ xy\equal{}\frac{6}{5}yx$.

$ 10x\plus{}y\equal{}\frac{6}{5}(10y\plus{}x)\equal{}12y\plus{}\frac{6}{5}x$

$ \frac{44}{5}x\equal{}11y, x\equal{}5 y\equal{}4$

answer : 54

<New Problem>

What is the area of the triangle with side lengths 12, 16, and 20?

[quote="mz94"]this is easier. call the 2 digit number (once again) 10a+b. then the sum is a+b
10a+b=2(a+b). simplifying gives us 8a=b. the smallest number is therefore when a=1, b=8

18

new prob:

find the 2 digit number which is 120% the number that is formed when tis digits are reversed[/quote]
10a+b=1,2(10b+a)
8,8a=11b
4a=5b   a divisible by 5 and a<10, so a=5, b=4 Answer: 54

<New problem>
Find x:
$ x\equal{} \frac {1}{1*2}\plus{}\frac {1}{3*4}\plus{}\frac {1}{5*6}\plus{}...\plus{}\frac {1}{99*100}$

[quote="Bachukas"]

<New problem>
Find x:
$ x \equal{} \frac {1}{1*2} \plus{} \frac {1}{3*4} \plus{} \frac {1}{5*6} \plus{} ... \plus{} \frac {1}{99*100}$[/quote]

to attempt to solve this we will investigate the first few terms

$ \frac {1}{1*2} \equal{} \frac {1}{2}$
$ \frac {1}{1*2} \plus{} \frac {1}{3*4} \equal{} \frac {3}{4}$
$ \frac {1}{1*2} \plus{} \frac {1}{3*4} \plus{} \frac {1}{5*6} \equal{} \frac {5}{6}$

well! it would appear

$ \frac {1}{1*2} \plus{} \frac {1}{3*4} \plus{} ...\frac {1}{n(n \plus{} 1)} \equal{} \frac {n}{n \plus{} 1}$

so since n in this case is 99, our answer is $ \frac {99}{100}$

and fantasy yours is 12*16/2 (its a right triangle), which is 96

new prob

3 Germans, 4 British, and 5 Americans sit in a row of chairs in a meeting. if they insist on sitting as groups, with each nationality being one group, how many seating combinations are there?

These problems are getting a little too hard. I'll post an easier one, because I'd hate to get this locked.

Mr. Holmes buys a chair for $ \$60$. He tries to sell it for double the price, but when no one would buy it, he reduced that doubled price by $ 60\%$. However, sales tax is $ 7\%$. Including tax, what was the final price of the chair?

So, here's what you do.

$ 60 \cdot 2 \equal{} 120$

But since no one wanted to buy it at that price, we reduce it by 60%.

$ 120 \cdot 0.4 \equal{} 48$

Add tax.

$ 48 \cdot 1.07 \equal{} \boxed{51.36}$

[hide="Problem 9"]If the hypotenuse of a right triangle is 25, then what is the smallest possible leg, if the two legs must be whole-integer lengths?[/hide]

Ernie, I believe that problem is too hard. Do you have another?

meh, ill just solve it and post another problem.

ernie's - 7 and 24

<Problem 10>

Evaluate: $ (3^4)^\frac{5}{4}$

THe fourth root of 3^4 cancels out to become 3. we then find 3^5 which is [size=150]243[/size]

In the future, please post a second problem. Also, I suggest reading the rules in the first post.

New Problem:

[Taken from my blog] Compute the following without using a calculator: $ \frac{8!}{6!2!}$.

[quote="Cortana"]EDIT: meant to write "$-9$" sorry

Let $n$ be a positive value in the expression $3^{2n}+4\cdot 3^{n}-9=0.$ What is $n$?[/quote]

I'm going to guess that the intended equation was $3^{2n}+6 \cdot 3^{n}-9=9$.

[hide="Solution to Previous Problem"]
Since the smaller cube is inscribed in the sphere, opposite corners are touching.  Therefore, if the radius of the sphere is $r$, then the length of a side of this cube is $\frac{2r}{\sqrt{3}}=\frac{2r\sqrt{3}}{3}$ because the distance between the opposite corners is both the diameter of the sphere and the space diagonal of the cube.
Since the larger cube is circumscribing the sphere, the exact middles of all the faces are touching.  Therefore, if the radius of the sphere is $r$, then the length of a side of this cube is $2r$.
Since $V = \frac{4}{3}\pi r^3$, the ratio of the smaller volume to the larger volume is
\[\dfrac{\dfrac{4}{3} \pi \left(\frac{2r\sqrt{3}}{3}\right)^3}{\dfrac{4}{3} \pi (2r)^3}\\=\dfrac{\left(\frac{2r\sqrt{3}}{3}\right)^3}{(2r)^3}\\=\left(\dfrac{\frac{2r\sqrt{3}}{3}}{2r}\right)^3\\=\left(\dfrac{2r\sqrt{3}}{2r(3)}\right)^3\\=\left(\dfrac{\sqrt{3}}{3}\right)^3\\=\dfrac{\sqrt{27}}{27}\\=\dfrac{3\sqrt{3}}{27}=\boxed{\dfrac{\sqrt{3}}{9}}\]
Whew!  Too much LaTeX!  :)[/hide]
[hide="New Problem"]
Compute $\sqrt{90+\sqrt{90+\sqrt{90+...}}}$
(Source: Inspired by several MathCOUNTS and AoPS problems.)[/hide]

EDIT:  Changed the description of my solution.

[hide="Solution to Previous Problem"]Let $x$ be $\sqrt{90+\sqrt{90+\sqrt{90+\cdots}}}$

Then,

$x = \sqrt{90+x}$

$x^2 = 90+x$

$x^2-x-90 = 0$

$x = -9, 10$

But since $x$ is positive, the answer is $\fbox{10}$.[/hide]

[hide="New Problem"]How many numbers from $1$ to $2011$, inclusive, are divisible by $3$ and $7$, but not $11$?[/hide]

[hide="Solution"]The least common multiple of $7$ and $3$ is $21$. So there are $\frac{2011}{21} \rightarrow 96$ numbers under $2011$ divisible by $3$ and $7$. The least common multiple of $21$ and $11$ is $231$. By similar computation, there are $9$ numbers under $2011$ divisible by $21$ and $11$. Thus, our answer is $96-9 \rightarrow \boxed{87}$.[/hide]

[hide="Problem"] The number $29844775923028N4$ is divisible by $11$. What is $N^2$?[/hide]

[hide="Solution"]
If $ 29844775923028N4 $ is divisible by $11$, that means the sum of every other number is equal to the sum of every other number had we started with the second digit.
In our case,
$ 2 + 8 + 4 + 7 + 9 + 3 + 2 + N = 9 + 4 + 7 + 5 + 2 + 0 + 8 + 4$
$ 14 + 16 + 5 + N = 13 + 12 + 2 + 12$
$ 35 + N = 39$
$ N = 4$
Therefore $N^2 = \boxed {16}$.[/hide]
[hide="New Problem"]
Suppose $10>x>y>0$. What is the greatest possible value of $\frac{x+y}{x-y}$?[/hide]

EDIT:  Edited problem due to my inconsideration.

Choosing numbers $x,y$ such that $10<x<y<0$ is pretty difficult ;)

This problem does not make any sence...
0 can't be greater than 10.
Are you sure you dont mean -10?

[hide="Solution"] To find the maximum value, we want to minimize the denominator and maximize the denominator and we know that $x$ cannot equal $y$. The largest numerator occurs when $x=9$ and $y=8$. Also, $9-8 =1$ so that is also the smallest possible denominator. Our answer is $\frac{9+8}{9-8}=\boxed{17}$ [/hide]

[hide="NP"] Equilateral  $\bigtriangleup ABC$ has a perimeter equal to the perimeter of square $WXYZ$. The area of $\bigtriangleup ABC$ is $144+36\sqrt{3}$. The area of square $WXYZ$ can be expressed in the form $m+n\sqrt{p}$ where $p$ is a number not divisible by the square of any prime. What is $m+n+p$?[/hide]

The previous problem was replaced by the new one to keep the marathon going. :D

[quote="tjt1005"][hide="Solution"]
The area of an equilateral triangle with side length $a$ is $\frac{a^{2}\sqrt{3}}{4}$.
Hence, $\frac{a^{2}\sqrt{3}}{4} = 144 + 36\sqrt{3}$.
$a = \sqrt{192\sqrt{3} + 144}$.
This is one side of the equilateral triangle.
The perimeter of the equilateral triangle is $3\sqrt{192\sqrt{3} + 144}$.
The perimeter of the square is $3\sqrt{192\sqrt{3} + 144}$.
One side of the square is $\frac{3\sqrt{192\sqrt{3} + 144}}{4}$.
The area of the square is $81 + 108\sqrt{3}$.
Hence, $m + n + p = 81 + 108 + 3 = 192$.
[/hide]

[hide="New Problem"]
A sheet of paper is given to two scanners, one human and one computer. The computer scanner misses out every 20th digit while the human scanner misses every 11th digit. If 20112011 digits are printed on the sheet of paper, how many digits did the human miss more than the computer?
[/hide][/quote]
[hide="Answer #???"]$\left\lfloor\frac{20112011}{11}\right\rfloor-\left\lfloor\frac{20112011}{20}\right\rfloor=1828364-1005600=\boxed{822764}$
That 's a lot of digits the human has to check![/hide]

[hide="P #???"]What is The probability of picking a 8 when you write down the digits of the first 100 even numbers?[/hide]

[hide="Solution"]The 100'th even number is 202.

There will be no 8's in the hundreds place.

There will be $2*1*5$ eights in the 10's place.

In the ones place, there will be $2*10*1$

Add the products, $10+20$=$30.$ $\frac{30}{202}= \frac{15}{101} [/hide]

[hide="New Problem"]32 workers are building a new school. It takes 5 days for them to build a room on the bottom floor. The school is to have 5 stories, each story 30 rooms. For every story that they go higher, it takes them 50 % more time to build than the previous story How long does it take for them to build the school? [/hide]

[quote="GausssWill"][hide="Solution"]The 100'th even number is 202.

There will be no 8's in the hundreds place.

There will be $2*1*5$ eights in the 10's place.

In the ones place, there will be $2*10*1$

Add the products, $10+20$=$30.$ $\frac{30}{202}= \frac{15}{101} [/hide][/quote]

[hide="Actually"]The 100th even number is \[200=99 \times 2+2\] therefore the answer is \[\boxed{\frac{3}{20}}\][/hide]
[hide="Solution"]We see that our answer is

\[\sum_{n=1}^5 150(1.5^{n-1})\\
\Longrightarrow \frac{150(1-1.5^5)}{1-1.5}\\
=\frac{15825}8 \approx \boxed{1978 \: \text{days}}\]
[/hide]
[hide="Problem"]Find the product of the sum of the factors of 328239 and 23892 that are multiples of 3. [/hide]

[hide=New Problem]Factor $x^{2018}+x+1$[/hide]

[color=#f00]Wolfram|Alpha is unable to do this computation, meaning that this problem is not MOEMS level.  Combined with the fact that this is a 6 year revive, topic locked.[/color]


NCSSM!

...

...

well, time to make another thread about the ARML competition chain.

how do you think you'll do?

At ARML?  I think NC will pwn everyone [color=white][size=34]at the Georgia Site[/size][/color]

Individually, I predict that my score will be $ \geq6$; I'm aiming for 7.

I believe he means the state math contest, though at this time of the year I'm much more concerned about the USAMO  :wink: .  I predict that I will make the ARML team  :D .  I hope the coaches give consideration to putting some of the middle schoolers on the A team.

Young talent has further to go.

anyway uh for State Math Contest I will probably not try to repeat my epic fail last year... which hey I didn't already repeat because I made it this year :wink:

but yeah top 5? I hope?

also USAMO is way more important than State Math Contest right now... and my mom is currently trying to get me to do state problems, but maybe I should go improve my olympiad combo instead?

edit: and I won't be going to ARML this year, so they will probably be forced to consider some younger students

Heh I was about to get really confused about why someone with a random username is acting like he knows what's going on, and then I realized who you are. Very tricky.

I would like to improve on last year :lol:. Really though I can do arbitrarily badly within the top 10 if I make mosp and arbitrarily badly in general if I win the usamo...

[quote="n^4+4"]I think this year should be a year all the future ARML folks get to know each other.[/quote]

What do you mean by this?

Umm...I am going to have to win State Math Contest to account for my failure at AMC/AIME.  Just saying.

I expect top 50. I hope top 30. (hopefully 25-ish)

[quote="MysticTerminator"][quote="n^4+4"]I think this year should be a year all the future ARML folks get to know each other.  We aren't going to win anything anyways.  Plus I wouldn't be surprised if the middle school students do better than some of the high school A team members.

Blargh in other news I lost $ \$$1000 in the final round of the itest ToC thing by 4 minutes :([/quote]

wtf[/quote]

This has to do with comprehensive/ARML how?

[quote="n^4+4"]I think this year should be a year all the future ARML folks get to know each other. We aren't going to win anything anyways. Plus I wouldn't be surprised if the middle school students do better than some of the high school A team members. [/quote]

[quote="b-flat"][quote="n^4+4"]We aren't going to win anything anyways.  [/quote][/quote]

How negative  :(

[quote="n^4+4"]Haha well my standards are kind of high after winning ARML two years ago...[/quote]

Just tell everyone on the 17th to do 33333333333333333 practice tests XD

[quote="Teki-Teki"]Another notice that has nothing to do with comprehensive/ARML.

PI-Dimension, remove the space after http: in your sig.

Also, I hope I don't get 6th again next year in Comprehensive.  That would be pretty inauspicious, considering I got 6th last year as well.[/quote]

There didn't put a space, its in LaTeX, which added a space I think. I tried fixing it, but it wouldn't work. The space just kept on being there.


first let's take out a common factor

$ 32x^3\minus{}4x^2\plus{}20x$

$ 4x(8x^2\minus{}x\plus{}5)$

and i believe that is as simplified as it gets

Actually $ 8x^2 \minus{} x \plus{} 5 \equal{} [x \minus{} \frac {1}{16}(1 \minus{} i\sqrt {159})][x \minus{} \frac {1}{16}(1 \plus{} i\sqrt {159})]$

Maybe they just wanted it factored with integer coefficients, maybe not..


abstract algebra
algebra
function
domain
calculus
integration
group theory

Let $ R$ be a noetherian integral domain whose field of fractions $ K$ is perfect and has non-zero characteristic. Prove that $ R=K$.

we only have to show $ R^{p}\equal{}R$. this is clear if $ R$ is factorial. no idea about the general case :-).

According to my sources, any noetherian local domain with fraction field K is dominated by a discrete valuation ring with the same fraction field. If we assume that $ R\neq K$, then $ R$ has a maximal ideal $ m$. Replacing $ R$ by $ R_m$ does not change the conclusion for $ R$, so we can assume that $ R$ is a noetherian local domain.

If $ (R,m)$ is dominated by a DVR $ (D,n)$ i.e. $ R\subseteq D$ AND $ m\subseteq n$, then we have on $ K$ a nontrivial discrete ($ \mathbb Z$ valued) valuation $ v$ induced by $ D$. Using that $ K$ is perfect thus closed under taking $ p$-th roots one can prove with ease that the image of $ v$ as a subgroup of $ Z$ must be a $ p$-divisible group. This is only the case for the $ 0$ subgroup which contradicts that $ v$ is nontrivial or if you don't like proofs by contradiction proves that every element of $ R$ is a unit in $ D$ thus $ R \equal{} D \equal{} K$ and $ n \equal{} 0$.

An algebraic geometric idea for proving that there exists a DVR that dominates $ R$ should go like this:
Consider $ X \equal{} Spec R$. Remember that $ R$ is a noetherian local domain. Blow-up the closed point of $ X$ corresponding to $ m$. This gives a scheme $ \tilde X$ that is birational to $ X$ so it has the same field of rational functions as $ X$ i.e. $ K$. The local ring at the generic point of the exceptional divisor of the blow-up should be our desired DVR.
I'd have to check this! I don't see why that local ring is not just of dimension 1.


function
real analysis
real analysis unsolved

Having a function $f: [a,b]\to\mathbb{R}$ show that if exists $c\in(a,b)$ such that $f$ has infinite limits in $c$ then $f$ is not monotonic on $[a,b]$

Assume that $f(x)$ tends towards $+\infty$ at $c$. Choose $a<d<c<e<b$ real numbers and $M>f(d)$, $M>f(e)$. Then you can find $g \in (d,e)$ such that $f(g)>M$. Hence $f$ is not monotonic on $(a,b)$.


calculus
integration
calculus computations

how would one evaluate
lim(n->oo) of
1/(n+1) + 1/(n+2) +...+ 1/(3n)
i know this is basically the sum from 1 to 2n of
1/(n+k) with respect to k as n approaches infinity
how would I do this? this appeared in an integral section of a math analysis problem book, but I do not know how to use integrals to evaluate such problems, help would be much appreciated.

$\frac1{n+k}=\frac1n\cdot\frac1{1+\frac{k}{n}}$

Can you see how to make this into a Riemann sum now?

i have idea but im not certain, i suppose 1/n would become the differential, but im rele uncertain on what to rewrite the integral as...i've never taken mathematical analysis, i'm teaching it to myself


logarithms
combinatorics unsolved
combinatorics

Prove that it is possible to colour any distinct $n$ points of the plane with at most $100.\ln n $colours in such a way that every circle containing at least one point contains exactly one of the points of some particular colour.

There is something that I do not understand : Take $n$ sufficiently large so that $n > 100 \times ln(n)$. Then at least two of the $n$ points will receive the same color. Now choose a circle passing through only these two points....

Pierre.

I guess it's meant to be discs, not circles.

it seems unclear to me too, but I thought the it's due to not knowing english well so I post it any way it's its URL:
http://www.komal.hu/verseny/2003-02/mat.e.shtml


请问谁知道米尔朵夫那本关于不等式的书哪里能免费下载到啊?(韩国人那本也行)
谢谢! :?:

学术类书籍一般不会有免费下载的吧，尤其是这本书本来就没什么名气的话


[size=150]1. Suppose U, V ⊂ Rⁿ are subspaces and U ⊂ V. Prove that V┴ ⊂ U┴.

2. Let V ⊂ Rⁿ be a subspace. Prove that V ⊂ (V┴)┴. Do you think more is true?[/size]

Note: V┴ is the orthogonal complement of V.

The wikipedia article [url]http://en.wikipedia.org/wiki/Orthogonal_complement[/url] said that that W┴┴ = W, but Question 2 says that V ⊂ (V┴)┴.  Do they contradict each other?

$ (V^\perp)^\perp\equal{}V$ when $ V$ is finite-dimensional, so wikipedia is probably assuming that. Also, note that it is true in your case since $ V\subset\mathbb{R}^n$. In general, you don't get equality though.

1. $ U \subset V \to V^ \perp \subset U^\perp$:

$ V^\perp \equal{} \{w\in R^n  |   <w,v>\equal{}0,  \forall v\in V\}$
$ U^\perp \equal{} \{w\in R^n  |   <w,u>\equal{}0,  \forall u\in U\}$

the usual subset argument works here.

2.$ V \subset \left(V ^ \perp\right)^\perp$:

$ \left(V ^ \perp\right)^\perp \equal{} \{w\in R^n  |   <w,v>\equal{}0,  \forall v\in V ^ \perp\}$
$ \equal{}\{w\in R^n  |   <w,v>\equal{}0,  \forall v$ with $ <v,u>\equal{}0,  \forall u\in V \}$

subset argument works again.

they are actually equal, so wikipedia isnt lying to you.


geometry
circumcircle
angle bisector
geometry unsolved

In a triangle $ ABC$, the angle bisector of $ \angle BAC$ intersects $ BC$ at $ D$. The tangents at $ D$ to the circumcircles of
the triangles $ ABD$ and $ ACD$ meet $ AC$ and $ AB$ at $ N$ and $ M$, respectively.Prove that the quadrilateral $ AMDN$ is
inscribed in a circle tangent to $ BC$.

[color=darkblue]We have $ \left\| \begin{array}{cc} \angle{ADM} \equal{} \angle{C} \\
\angle{NDA} \equal{} \angle{B} \end{array} \right\|$, so $ \angle{MAN} \plus{} \angle{MDN} \equal{} \angle{A} \plus{} \angle{B} \plus{} \angle{C} \equal{} 180^\circ$, so quadrilateral $ AMDN$ is cyclic. We have $ \angle{C} \equal{} \angle{ADM} \equal{} \angle{MNA}$$ \Rightarrow MN \parallel BC \Rightarrow \angle{MND} \equal{} \angle{MAD} \equal{} \angle{DAN} \equal{} \angle{DMN} \equal{} \angle{MDB} \Rightarrow BC$ is tangent to the circumcircle of $ AMDN$.[/color]


AMC
AMC 12
AMC 12 A
AMC 12 B
AIME
USA(J)MO
USAMO

im just wondering

is cutoff for amc12A and B usually similar?

or is B lower

or is B higher?

No one knows until they grade the exams...

i would assume them to be somewhat similar... it would be unfair if one AMC version was harder than the other, although that does sometimes inadvertently happen

You can find the answer to your question by consulting the general statistics on the AMC 12 Archive:
http://www.unl.edu/amc/e-exams/e6-amc12/archive12.shtml

Steve Dunbar
Director, American Mathematics Competitions

[quote="yc"]AMC A and B  AIME qualifying scores

2002:   10 A:115. 10 B: 118.         12A and 12B: 100

2003:    10A: 119. 10B: 121.         12A and 12B: 100

2004:    10A: 110. 10B: 120.         12A AND 12B : 100

2005 :   10A  AND 10B: 120 .        12A AND 12B: 100

2006:   10A AND 10B: 120.           12A AND 12B: 100.[/quote]

I could have sworn that we determined that the 12A cutoff was 84 in 2002 in one of the other threads...

Anyway, the only thing that really matters is how they calculate the index.  Do they do that separately for the A and B?

I dont think the cutoff has ever been anything different than 100 except this years A (or possibly B, from what I heard), so they're usually both 100.
However, this year, it might be different.

[quote="DanK"]I could have sworn that we determined that the 12A cutoff was 84 in 2002 in one of the other threads...

Anyway, the only thing that really matters is how they calculate the index.  Do they do that separately for the A and B?[/quote]

The cutoff that was 84 was on the 2001 test.

I don't think they've ever made a distinction for USAMO between the AMC 12 A and B, but they have for AIME I and II.  But that's only if the tests are really different in difficulty.


[quote="mathmaniack"]Let ABC be any triangle and d',d'',d''' distance from orthocentre to the sides of triangle.If  r is incircle radius show that  d'+d''+d''' <=3.r[/quote]

First, the triangle ABC should be acute-angled, else this is wrong. Then, see http://www.mathlinks.ro/Forum/viewtopic.php?t=47770 .

  darij


Let $ x,y,z\geq 0$ such that $ : x^2\plus{}y^2\plus{}z^2\equal{}2$.Prove that $ : \sum\frac{x^2}{y\plus{}z}\leq\frac{3}{2}$

[quote="DCTPKTCSPHN"]Let $ x,y,z\geq 0$ such that $ : x^2 \plus{} y^2 \plus{} z^2 \equal{} 2$.Prove that $ : \sum\frac {x^2}{y \plus{} z}\leq\frac {3}{2}$[/quote]

Are you sure? $ y\plus{}z$ can be made arbitrarily small and so $ L.H.S.$ can be made arbitrarily large.

[quote="DCTPKTCSPHN"]Let $ x,y,z\geq 0$ such that $ : x^2 \plus{} y^2 \plus{} z^2 \equal{} 2$.Prove that $ : \sum\frac {x^2}{y \plus{} z}\leq\frac {3}{2}$[/quote]

Check:
$ x\equal{}y\equal{}1;z\equal{}0$

[quote="stephencheng"][quote="DCTPKTCSPHN"]Let $ x,y,z\geq 0$ such that $ : x^2 \plus{} y^2 \plus{} z^2 \equal{} 2$.Prove that $ : \sum\frac {x^2}{y \plus{} z}\leq\frac {3}{2}$[/quote]

Are you sure? $ y \plus{} z$ can be made arbitrarily small and so $ L.H.S.$ can be made arbitrarily large.[/quote]

I agree. Also, what's the summation over on the right hand side?

[quote="DCTPKTCSPHN"]Let $ x,y,z\geq 0$ such that $ : x^2 \plus{} y^2 \plus{} z^2 \equal{} 2$.Prove that $ : \sum\frac {x^2}{y \plus{} z}\leq\frac {3}{2}$[/quote]
Sorry,I want say
Let $ x,y,z\geq 0$ such that $ : x^2 \plus{} y^2 \plus{} z^2 \equal{} 2$.Prove that $ : \sum\frac {x^2}{x\plus{} y}\leq\frac {3}{2}$


A large for-profit corporation bought a controlling interest in a furniture manufacturer and changed the latter’s policies. The furniture manufacturer lost money, but the corporate executives were pleased. Why?

[hide="Because..."]They were now buying all of their furniture at below cost.[/hide]


[quote="zaya_yc"]
 $ D \equal{} ( \minus{} 4a^{3})^{2} \minus{} 4*2a^{5}*(1 \plus{} a^{2}) \equal{} \minus{} 8a^{5}*(1 \minus{} a)^{2}\leq 0$  and $ a > 0$ ---> $ D \equal{} 0$   ----> $ a \equal{} 2$ --> $ x \equal{} 1$[/quote] doesn't work...review your calculus...here's my proof: [hide]$ \frac {9^x \plus{} 36^x \plus{} 32^x \plus{} 32^x}{4} \equal{} 24^x \geq \sqrt [4]{(9 \cdot 36 \cdot 32 \cdot 32)^x} \equal{} 3^x \cdot 2^{3x} \equal{} 24^x$ and this is the AM-GM equality case, therefore $ 9^x \equal{} 36^x \equal{} 32^x$ and $ x \equal{} 0$[/hide]

[quote="zaya_yc"][quote="balan razvan"]Solve the equation
\[ 9^x \plus{} 36^x \plus{} 2\cdot 32^x \equal{} 4 \cdot 24^x
\]
I hope this equation was not posted before :)[/quote]

$ 2^{x} \equal{} a$ , $ 3^{x} \equal{} b$ --->  $ b^{2} \plus{} a^{2}b^{2} \plus{} 2a^{5} \equal{} 4a^{3}b$ --->

$ (1 \plus{} a^{2})b^{2} \minus{} 4a^{3}b \plus{} 2a^{5} \equal{} 0$

$ D \equal{} ( \minus{} 4a^{3})^{2} \minus{} 4*2a^{5}*(1 \plus{} a^{2}) \equal{} \minus{} 8a^{5}*(1 \minus{} a)^{2}\leq 0$  and $ a > 0$ ---> $ D \equal{} 0$   ----> $ a \equal{} 2$ --> $ x \equal{} 1$[/quote]
 the only mistake is that the equality fot the $ \minus{} 8a^{5}*(1 \minus{} a)^{2}\leq 0$stands only if a=1 so x=0 :wink:


We know that a continuous bijective function from a compact space to a Hausdorff space must be a homeomorphism. However, I have the following questions in mind:

1. Give an example where the map is not a homeomorphism if the target space is not Hausdorff.
2. Must a continuous bijective function from a Hausdorff space to a compact space be a homeomorphism?

Thank you very much for your help!!

1.  Let the target space have the indiscrete topology with more than one point.  No continuous bijection can be an open map.

2.  No.  Perhaps the best explanation I have seen for why this exercise is true is [url=http://terrytao.wordpress.com/2009/02/09/245b-notes-10-compactness-in-topological-spaces/]this post of Terence Tao's[/url].


I'm dumb at conics, what is eccentricity???

[quote="shobber"]Through the left focus $F_1$ of $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ construct a line intersecting the ellipse at $A$ and $B$. If $\triangle{ABF_2}$ is a isocele right triangle with $\angle{A}=90^o$. ($F_2$ is the right focus). Find the eccentricity of the ellipse.[/quote]

[hide]$ABF_2$ is a right angle isosceles triangle,if $AF_1=m$ and $BF_1=n$ then: $AB=AF_2=m+n$ and $BF_2=\sqrt{AB^2+AF_2^2}=\sqrt{2}(m+n)$

Since $F_1$ and $F_2$ are the focus of the ellipse :$AF_1+AF_2=BF_1+BF_2 \implies 2m+n=\sqrt{2}m+(\sqrt{2}+1)n$
$\implies n=(\sqrt{2}-1)m$

but $2a=AF_1+AF_2=(\sqrt{2}+1)m$ and $F_1F_2=\sqrt{AF_1^2+AF_2^2}=\sqrt{3}m$,the eccentricity is $\frac{F_1F_2}{2a}=\frac{\sqrt{3}m}{(\sqrt{2}+1)m}=\sqrt{6}-\sqrt{3}$ [/hide]

You are correct, Gabriel(fr).

pkerichang, eccentricity follows the definition of the conic. It is the ratio of the distance between the points and the distance between the point and the directrix.


function
AMC
USA(J)MO
USAMO
inequalities unsolved
inequalities

Show that for all non-negative reals $ a,b,c\leq 1$
$ \sum_{cyc}\left[\frac{a}{b \plus{} c \plus{} 1}\right] \plus{} (1 \minus{} a)(1 \minus{} b)(1 \minus{} c)\leq 1$

[quote="cnyd"]Show that for all non-negative reals $ a,b,c\leq 1$
$ \sum_{cyc}\left[\frac {a}{b \plus{} c \plus{} 1}\right] \plus{} (1 \minus{} a)(1 \minus{} b)(1 \minus{} c)\leq 1$[/quote]
Let $ f(a,b,c)\equal{}\sum_{cyc}\frac {a}{b \plus{} c \plus{} 1} \plus{} (1 \minus{} a)(1 \minus{} b)(1 \minus{} c).$
Hence, $ f$ is a convex function of $ a,$ of $ b$ and of $ c.$ Thus,
$ \max_{\{a,b,c\}\subset[0,1]}f\equal{}\max\{f(0,0,0),f(0,0,1),f(0,1,1),f(1,1,1)\}\equal{}1.$

[quote="keyree10"]What does it mean, to call a multi-variable function, convex?[/quote]
I have meant that if $ g(a)\equal{}f(a,b,c),$ $ h(b)\equal{}f(a,b,c)$ and $ p(c)\equal{}f(a,b,c)$
then, $ g,$ $ h$ and $ p$ are convex functions on $ [0,1].$ :wink:

So, it should be a convex function with respect to each variable, treating the other variables like constants? And just like a one-variable convex function, this also attains maximum value at the endpoints?

Yes keyree10, it is actually the Theorem of Weirstrass that a conex function attains maximum at end points.
You get by double differentiating.

[quote="cnyd"]Show that for all non-negative reals $ a,b,c\leq 1$
$ \sum_{cyc}\frac {a}{b \plus{} c \plus{} 1} \plus{} (1 \minus{} a)(1 \minus{} b)(1 \minus{} c)\leq 1$[/quote]

without loss of generality, we can assume that $ 0\le a \le b \le c \le 1$, then we have
$ \sum_{cyc}\frac {a}{b \plus{} c \plus{} 1} \plus{} (1 \minus{} a)(1 \minus{} b)(1 \minus{} c)$
$ \le \frac {a \plus{} b \plus{} c}{a \plus{} b \plus{} 1} \plus{} (1 \minus{} a)(1 \minus{} b)(1 \minus{} c)$
$ \equal{} 1 \minus{} \frac {1 \minus{} c}{a \plus{} b \plus{} 1}\left(1 \minus{} (a \plus{} b \plus{} 1)(1 \minus{} a)(1 \minus{} b)\right)$
$ \le 1 \minus{} \frac {1 \minus{} c}{a \plus{} b \plus{} 1}\left(1 \minus{} (ab \plus{} a \plus{} b \plus{} 1)(1 \minus{} a)(1 \minus{} b)\right)$
$ \equal{} 1 \minus{} \frac {1 \minus{} c}{a \plus{} b \plus{} 1}\left(1 \minus{} (1 \minus{} a^2)(1 \minus{} b^2)\right)$
$ \le 1$

To be honest, my solution is exactly same as yours. :lol:

[quote="cnyd"]Show that for all non-negative reals $ a,b,c\leq 1$
$ \sum_{cyc}\left[\frac {a}{b \plus{} c \plus{} 1}\right] \plus{} (1 \minus{} a)(1 \minus{} b)(1 \minus{} c)\leq 1$[/quote]

How about this one with the same conditon???
$ \sum_{cyc}\left[\frac {a}{b \plus{} c \plus{} 1}\right] \plus{} (1 \minus{} a)(1 \minus{} b)(1 \minus{} c)\geq \frac {7}{8}$

Is it hard? :maybe:


geometry
parallelogram
geometry proposed

[I do not know if this is already posted, but is nice and not so easy :)]

Let ABC be a triangle and $D \in (BC)$, $E \in (CA)$, $F \in (AB)$ such that $BD = CE = AF$ and $m(BAD) = m(CBE) = m(ACF)$,
where $m(XYZ)$ is the measure of the angle XYZ.
Prove that ABC is equilater.

Let G be the point so that AFCG is a parallelogram, then, since ADG ~ ABC, also EBDG is a parallelogram. So ECG is equilateral then we are gone.

Edit.

I'm sorry but it is wrong. I have to think about more :-).

Any progress on this one? I tried some computations, but I soon reached a dead end.


geometry
circumcircle
Euler
power of a point
radical axis
geometry proposed

Suppose $O$ is circumcenter of triangle $ABC$. Suppose $\frac{S(OAB)+S(OAC)}2=S(OBC)$. Prove that the distance of $O$ (circumcenter) from the radical axis of the circumcircle and the 9-point circle is \[\frac {a^2}{\sqrt{9R^2-(a^2+b^2+c^2)}}\]

I suppose the notation $S(OAB)$ is the same as $S_{\triangle OAB}$ or $[OAB]$ :?
I haven't gave a thought in it, but I believe it can be solved by using the formula $OH^2=9R^2-(a^2+b^2+c^2)$.

$S_{ABC}=3S_{BOC}(1)$let $AO$ intersect $BC$ at $A'$ from $(1)$ we can get$\frac {AO}{OA'}=2$ so Euler line is parallel to $BC$.
in triangle $AOH$ use pythagores theorem and $OH^2=9R^2-(a^2+b^2+c^2)$ formula .
with Casi theorem  we can calculate the distancse

If it is Casey Theorem, then you can find it in the Geometry Formula section.

the statement is wrong, if $\triangle ABC$ is acute, then the distance should be $\[\frac {a^2}{4\sqrt{9R^2-(a^2+b^2+c^2)}}\]$. and I believe that the result will be different when $\triangle ABC$ is obtuse.


I am new to linear algebra and im just reading a book and studying on my own for the next semester. One of my friends gave me a sample exam on his linear algebra class a year ago.

Here it goes.

Consider the systems below (3 x 5 matrix but dont konw how to latex it)

$ 2x_1 \plus{} 4x_5 \equal{} 3000$
$ 2x_2 \plus{} 4x_3 \equal{} 8000$
$ x_2 \minus{} 2x_4 \plus{} 3x_5 \equal{} 5000$

1) Find the canonical form with respect to $ x_3$, $ x_4$ and $ x_5$?
 (What on earth is canonical form?)
2.) What is the basic solution to this canonical form?
3.) Find 3 non-basic solutions of the system.
(What's a non-basic solution?)

It's not standard terminology as I know it. Find the text your friend was using and look for definitions there.


calculus derivative inequalities algebra polynomial integration trigonometry

Let $P(x) = a_nx^n + ...... a_0$ such that $|P(x)| \leq 1$ for every x in $[-1, 1]$. Prove that $|P'(x)| \leq n^2$

So what happens santos_blr ? The problem is not easy as you think. :D

Clear this up for me: do you mean $|P'(x)|\le n^2,\ \forall x\in[-1,1]$?

This gave me headaches today. I spent two hours on it during the math course( I know I shouldn't have said that) and I could not solve it. I think it is far from a simple problem and the result is more than beautiful. Moubi, what you wrote there is the general form of Markov inequality? I think flip2004 also wrote something like this, but I do not remember clearly. Anyway, nice problem. And yes, grobber, this is what we are supposed to prove. Quite hard, isn't it?

Here a version for polynomial trigonometric,
it was an oral examination Ecole Polytechnique

1/ $f(x)=\sum_ {k=0}^{n}(a_k.cos(kx) +b_k.sin(kx)) $ polynomial trigonometric
of degree $\leq n$
We suppose there exist $x_1,x_2,...,x_{2n+1}$ distincts reals in $[0;2\pi[$
Prove that f=0
2/Let f a polynomial trigonometric  $f(x)=\sum_ {k=0}^{n}(a_k.cos(kx) +b_k.sin(kx)) $ of degree $\leq  n$ such that $f'(0)=||f'||>n||f||$  where $||f||=sup|f|$
Let $g(x)=1/n.||f'||sin(nx) - f(x)$
Prove that g vanishes at $2n$ reals in $[0;2\pi]$, and g' vanishes at least
2n+1 reals in $[0;2\pi]$
Deduce that $g\"=0$ and get a contradiction.
3/Prove that $||f'||\leq n||f||$

Oops, I 'm sorry. It should be $f'(x) \leq n^2 $ for every x in $[-1.1].
 Moubinool 's result is right too by applying this a few times.

moubinool's result is much stronger than just applying your inequality several times.
unfortunately, i do not see any way to prove that beautiful inequality. i guess i'll be spending my two free lessons tomorrow trying to prove it.

Peter

This follows easily from elementary theory of moment problem, check out "The Classical Moment Problem" by Akhiezer, it's much more general than this and there are generalizations for all positive linear functionals (in this case "integrals" with orthogonal polynomials).

Here is a more elementary proof outlined:

(1) Prove that all polynomials $P$ with $\deg(P) = n - 1$ which satisfy $|P(x)| \leq 1/\sqrt(1-x^2)$ for $x \in [-1,1]$ also satisfy $|P(x)| \leq$ for $x \in [-1,1]$ with equality if $P(x) = c U_{n-1}(x)$ for some suitable constraints on $c$ where $U_{n-1} = T'_n(x)/n$ and T are the Chebychef polynomials.

(2) Let $P(x) = \sum_{k=1}^n \lambda_k \sin(kx)$ with real coefficients for which $P(x) \leq$ holds then $|P(x)/\sin(x)| \leq n$.

(3) Let $Q(x) = \lambda_{-1} + \lambda_0\cos(x) + \mu_1\sin(x) + \cdots + \lambda_n\cos(nx) + \mu_n\sin(nx)$ with for all real $x$ $|Q(x)| \leq 1$. Show that $|Q'(x)| \leq n$, consider $P(x) = (Q(x_0 + x) - Q(x_0 - x))/2$.

(4) Now show the problem given here by looking at $P(\cos(x))$.

[quote="Moubinool"]Here a version for polynomial trigonometric,
it was an oral examination Ecole Polytechnique

1/ $f(x)=\sum_ {k=0}^{n}(a_k.cos(kx) +b_k.sin(kx))$ polynomial trigonometric
of degree $\leq n$
We suppose there exist $x_1,x_2,...,x_{2n+1}$ distincts reals in $[0;2\pi[$
Prove that f=0
2/Let f a polynomial trigonometric  $f(x)=\sum_ {k=0}^{n}(a_k.cos(kx) +b_k.sin(kx))$ of degree $\leq n$ such that $f'(0)=||f'||>n||f||$  where $||f||=sup|f|$
Let $g(x)=1/n.||f'||sin(nx) - f(x)$
Prove that g vanishes at $2n$ reals in $[0;2\pi]$, and g' vanishes at least
2n+1 reals in $[0;2\pi]$
Deduce that $g\"=0$ and get a contradiction.
3/Prove that $||f'||\leq n||f||$[/quote]

I would like to know if you had any idea of how to demonstrate that linear combination of $\sin (nx)$ and$\cos (nx)$ are the only functions that realize $||f'||=n||f||$


if the exponential equation whose graph passes through the points (1,5) and (2,7) is given by $ y \equal{} a(b)^x$, find a

2. also, what is the equation of the asymptote of $ y\equal{}\minus{}13(\frac{1}{3})^x$

can you explain how to do these ?

1. Um... plug points in? System of equations.

2. Well, it's defined for all $ x$, so no vertical asymptotes here; you're looking for a horizontal asymptote. What's $ \lim_{x\to\infty}\minus{}13\cdot\left(\frac{1}{3}\right)^x$?

hmm, im pretty sure there is an easier method for both of them since the answer choices are pretty simple...

for 1, the choices are 7/5, 5/7, 25/7, 1/5, NOTA

and for 2, y=-13, y=0, y=x, y=1/3

Well, for number 2, since $ \lim_{x\rightarrow \infty}\minus{}13\left( \frac13 \right)^x \equal{} 0$, your answer is $ y\equal{}0$.

Number one is quite straightforward, just do as grn_trtle says.

Where are these problems from, anyway?

For 1, plugging in points gives $ 5\equal{}ab$ and $ 7\equal{}ab^2$. From this finding $ a$ should be no trouble at all.

For 2, that's as simple as it gets. If it makes it easier, $ \lim_{x\to\infty}\minus{}13\cdot\left(\frac{1}{3}\right)^x\equal{}\minus{}13\cdot \lim_{x\to\infty}\frac{1}{3^x}$. (Or just think about what happens when $ x$ gets really huge).

Edit: Yes, DaReaper is correcct on #2.

these are hw probs.

for number two, im certain that there is an easier way b/c we havnt learned limits yet, and this was on the hw... (maybe by graphing, how would you graph this)

Please mention that they're homework problems when you post them, so that people don't think you're cheating.

Whether you know limits or not, the idea behind horizontal asymptotes is to see what $ y$ converges to as $ x$ goes in some direction. So yes, you could sketch it by hand fairly easily (passes through $ (0,1)$, starts very negative and gets less negative), and then it's pretty clear that it converges to $ 0$, or you can just think about what happens when $ x$ gets very large (the fraction gets really small, and the whole thing gets close to $ 0$).

Ok for these problems, you know that a^x (in this case a is 1/3) can not be 0 or any negative number. So since 0 is the closest to any possible a^x, input (1/3)^x as 0 and see how everything else plays out.

Since -13*0=0, the asymptote is 0.

Btw how do you define a limit?

Well, $ \lim_{x \to L}{f(x)}$ denotes the value that $ x$ [i]approaches[/i] as $ x$ gets closer to $ L$. A lot of the time, this turns out to be simply $ f(L)$. When $ L$ is replaced with $ \infty$ or $ \minus{}\infty$, then we are looking for the value that $ f(x)$ converges to as it gets larger and larger, or smaller and smaller. If $ f(x)$ diverges, which means it increases or decreases without bound (or oscillates), then the limit is simply $ \infty$ or $ \minus{}\infty$. You can take a look in a precalculus or elementary calculus book if want to learn more, or [url=http://en.wikipedia.org/wiki/Limit_(mathematics)]go here[/url].


The product of uncommon real roots of the two polynomials $ x^4 \plus{} 2x^3 \minus{} 8x^2 \minus{} 6x \plus{} 15$ and $ x^3 \plus{} 4x^2 \minus{} x \minus{} 10$ is ?

$\textbf{(A)}\  \minus{} 4 \qquad\textbf{(B)}\  4 \qquad\textbf{(C)}\  \minus{} 6 \qquad\textbf{(D)}\  6 \qquad\textbf{(E)}\ \text{None}$

[hide="Let me tell you how i solved the problem"]
$ x \equal{} \minus{}2$ is a root of the second polynomial $ Q(x)$. So $ Q(x) \equal{} (x\plus{}2)(x^2\plus{}2x\minus{}5)$.

$ x \equal{} \minus{}2$ is not a root of the first polynomial $ P(x)$. Then check if $ (x^2\plus{}2x\minus{}5)$ is the root of $ P(x)$.

$ P(x) \equal{}(x^2\minus{}3)(x^2\plus{}2x\minus{}5)$. So the product of uncommon real roots is $ \minus{}3 \cdot \minus{}2 \equal{} 6$.

$ \boxed D$

[/hide]


geometry
trigonometry
ratio
circumcircle
geometry unsolved

In triangle $ ABC$ we have $ AB\equal{}AC$.The bisector of angle $ \angle ABC$ meets $ AC$ in $ D$.It is known that $ BC\equal{}2AD\plus{}AB$.
Find the angles of the triangle.

$ BC \equal{} 2AD \plus{} AB \ \Longleftrightarrow \ a \equal{} \frac{2c^2}{c\plus{}a} \plus{}c  \ \Longleftrightarrow \ c\equal{}\frac{a}{\sqrt{3}}$

Then $ \sin (\angle ABC) \equal{} \frac{\sqrt{3}}{2} \ \Longrightarrow \ \angle ABC \equal{} \frac{\pi}{6}$

Here is my triginometric solution.
$ BC \equal{} 2AD \plus{} AB \equal{} 2AD \plus{} AC \equal{} 3AD \plus{} CD$.
By bisector theorem we obtain that $ AD \equal{} \frac {AB\cdot CD}{BC} \plus{} CD$. Then
$ BC \equal{} 3\frac {AB\cdot CD}{BC} \plus{} CD \plus{} CD$. Divide by $ CD$.
$ \frac {BC}{CD} \equal{} 3\frac {AB}{BC} \plus{} 1$.         * [b]$ (1)$[/b]
Let $ \angle{B} \equal{} \angle{C} \equal{} 2x, \angle{A} \equal{} 180^\circ \minus{} 4x, \angle{BDC} \equal{} 180^\circ \minus{} 3x$.
In $ \triangle{BDC}$ we have: $ \frac {BC}{CD} \equal{} \frac {\sin{3x}}{\sin{x}} \equal{} \frac {\sin{x}(3 \minus{} 4\sin^2{x})}{\sin{x}} \equal{} 1 \plus{} 2(1 \minus{} 2\sin^2{x}) \equal{} 1 \plus{} 2\cos{2x}$
In $ \triangle{ABC}$ we have: $ \frac {AB}{BC} \equal{} \frac {\sin{2x}}{\sin{4x}} \equal{} \frac {1}{2\cos{2x}}$
By replacing this ratios in $ *(1)$ we have:
$ (1 \plus{} 2\cos{2x})2\cos{2x} \equal{} 3 \plus{} 2\cos{2x}$
$ 2\cos{2x} \plus{} 4\cos^2{2x} \equal{} 3 \plus{} 2\cos{2x}$
$ \cos^2{2x} \equal{} \frac {3}{4}$. $ \cos{2x} \equal{} \frac {\sqrt {3}}{2}$, so $ 2x \equal{} 30^\circ$

$ \angle{ABC} \equal{} 30^\circ$

[hide="Solution"]
Construct points $ E$ and $ F$ on the extension of $ BA$ such that $ AE \equal{} AD \equal{} EF$ and $ AF \equal{} 2AD$.  Furthermore, let $ BD$ meet $ FC$ at $ K$, let the orthocenter of $ \triangle ADF$ be $ H$ and let its circumcenter be $ O$.  Now, we see that $ BC \equal{} AB \plus{} 2AD \equal{} AB \plus{} AF \equal{} FB$.  Let $ \angle ABD \equal{} \angle DBC \equal{} \alpha$ and thus $ \angle ABC \equal{} 2\alpha\implies \angle ACB \equal{} 2\alpha$.  Thus, $ \angle ADB \equal{} \angle DBC \plus{} \angle DCB \equal{} 3\alpha$ and $ \angle BAC \equal{} 180 \minus{} \angle ABC \minus{} \angle ACB \equal{} 180 \minus{} 4\alpha$.  Furthermore, $ BK$ is a bisector of $ \angle FBC$ and $ FB \equal{} BC$, so $ BK\perp FC$.  It follows that $ \angle KCB \equal{} 90 \minus{} \angle KBC \equal{} 90 \minus{} \alpha\implies \angle KCD \equal{} \angle KCB \minus{} \angle DCB \equal{} 90 \minus{} 3\alpha$.  $ rac {\angle BAC}{2} \equal{} 90 \minus{} 2\alpha$.  It follows that $ \angle EDF \equal{} 180 \minus{} \angle ADE \minus{} \angle ADB \minus{} \angle FDK \equal{} 90 \minus{} 4\alpha$.  We can see that $ \angle DAF \equal{} 180 \minus{} \angle DAF \equal{} 4\alpha$ and thus, $ \angle AFD \equal{} 180 \minus{} \angle FAD \minus{} \angle ADF \equal{} 2\alpha$.  We conclude that $ \angle ADH \equal{} 90 \minus{} \angle DAF \equal{} 90 \minus{} 4\alpha \equal{} \angle EDF$.  It is well-known that $ \angle FDO \equal{} \angle ADH$, which implies that $ O$ is on $ DE$.  Since $ O$ is the intersection between the perpindicular bisector of $ AF$ and $ DE$ and $ E$ lies on the perpindicular bisector of $ AF$ since $ AE \equal{} EF$, we see that $ E \equal{} O$, so $ \angle ADF \equal{} 90$.  This means that \[ 90 \equal{} \angle ADF \equal{} \angle ADE \plus{} \angle EDF \equal{} 90 \minus{} 4\alpha \plus{} 90 \minus{} 2\alpha \equal{} 180 \minus{} 6\alpha\implies \alpha \equal{} 15\implies \angle BAC \equal{} 180 \minus{} 4\alpha \equal{} 120\] and $ \angle ABC \equal{} \angle ACD \equal{} 2\alpha \equal{} 30$.  [/hide]

[quote="Virgil Nicula"][color=darkred][b][u]Remark.[/u][/b] Let $ ABC$ be a triangle. The bisector of $ \widehat {ABC}$ meets the side $ [AC]$ in the point $ D$.

It is known that $ BC \equal{} 2\cdot AD \plus{} AB$. Prove that $ \cos\frac A2\equal{}\frac {b}{2c}$ .[/color][/quote]

[color=darkblue][b][u]Proof.[/u][/b] Indeed, $ BC\equal{}2\cdot AD\plus{}AB$ $ \Longleftrightarrow$ $ a\equal{}2\cdot \frac {bc}{ a\plus{}c}\plus{}c$ $ \Longleftrightarrow$ $ a^2\minus{}c^2\equal{}2bc$ $ \Longleftrightarrow$

$ \left(b^2\plus{}c^2\minus{}2bc\cdot\cos A\right)\minus{}c^2\equal{}2bc$ $ \Longleftrightarrow$ $ b\equal{}2c(1\plus{}\cos A)$ $ \Longleftrightarrow$ $ b\equal{}4c\cos^2\frac A2$ $ \Longleftrightarrow$ $ \boxed {\ \cos\frac A2\equal{}\frac {b}{2c}\ }$ .[/color]


Gauss
algorithm
algebra
system of equations

[quote="debanik2"]solve the following for x,y,z :

10x - 20y = 4
20y - 30z = 1
30z - 10x = -5[/quote]

wrong forum

hello, simplifying your system we get
$ 5x\minus{}10y\equal{}2$
$ 20y\minus{}30z\equal{}1$
$ \minus{}2x\plus{}6z\equal{}\minus{}1$
and applying the Gauss-algorithm we get the solution
$ x\equal{}\frac{2}{5},y\equal{}0,z\equal{}\minus{}\frac{1}{30}$
Sonnhard.

There isn't a unique solution to this because equations (1) and (2) imply equation (3).

But we can assume $ x\equal{}k$ and, from (3), find $ z\equal{}\frac{2k\minus{}1}{6}$ and from (1) $ y\equal{}\frac k2\minus{}\frac 15$.

So the solutions are $ \boxed{(x,y,z)\equal{}\left(k,\frac{2k\minus{}1}{6},\frac k2\minus{}\frac 15\right)}$.

hello, you have right, it was a mistake of me.
Sonnhard.

Thank u all

Here (1) and (2) give (3)

are there any more such systems of equations where 3 is related to (1) and (2) by some means?

Is it right to conclude everytime that there will no unique soln for such systems of equations?

In a system of equations where there are more variables than equations there are infinitely many solutions (in the real numbers). This is the case here, because one of the equations doesn't add any new information.


geometry
3D geometry
tetrahedron
trigonometry

In the tetrahedron ABCD we have $ AB\equal{}BC\equal{}CA\equal{}4\sqrt{3}$, $ DC \equal{} 4\sqrt{2}$, $ M$ the midpoint of $ AB$ and $ DM \equal{} 2\sqrt{5}$. If $ DC$ is perpendicular to $ AB$ then compute the distance from $ D$ to the plane ABC, the sine and cosine of the angle formed by the plane $ ABC$ and $ ABD$ and the angle formed by CD with the plane $ ABC$.

Sorry, I thought it was hard, this is more like intermediate. I solved it!

If you want, I'll post the results, but I'll leave it a few days for you to think for a solution.


probability
geometry
3D geometry
tetrahedron
octahedron
dodecahedron
icosahedron

For the following five problems, imagine there is an ant at each vertex and that the ants all simultaneously crawl along an edge to the next vertex, each ant choosing its path randomly. What is the probability that no ant will encounter another, either en route or at the next vertex, for each of the following regular polyhedrons? (Express your answer as a reduced fraction; e.g., 2/8 = 1/4.)
38. 	A tetrahedron [cf. Hoeflin Power Test, problem 21]
39. 	A cube [cf. Hoeflin Power Test, problem 22]
40. 	An octahedron [cf. Hoeflin Power Test, problem 23]
41. 	A dodecahedron  [cf. Hoeflin Power Test, problem 24]
42. 	An icosahedron

What's the cleanest, most clever way to do this?

[quote="lotrgreengrapes7926"]Isn't this an ongoing IQ test? I don't think we should be discussing these questions.  :maybe:[/quote]

What IQ test?

I posted it because it was a pretty cool math problem I saw somewhere.

So, does anyone have a good answer that doesn't involve lots of case work? Have a few cases and rotating them, maybe?

Here's a re-write of the more general form of this problem:

"Given a graph G, compute the number of "directed subgraphs" of the graph composed of (directed) cycles of length greater than 2 such that each vertex is in exactly one cycle."

The key for the cube and dodecahedron case is to realize that the shortest possible cycles are 4 and 5, respectively, making the total number of cases to consider very small.  For the cube, for instance, there are only two cases: the cycle which covers all 8 vertices (there is only one up to rotations) and pairs of cycles, the first covering one face and the second covering the opposite face.  This is why the icosahedron case is difficult: there are many vertices, which allows for the possibility of many cycles, which means very many cases to consider.

What does it mean for an IQ test to be "ongoing?"

The IQ test is supposedly for really smart people. You find it on the web, take it over the course of a month, and send it in. It's a load of bunk, though. Most of the spatial reasoning section is either trivial or an easy casework puzzle. The verbal section is easily google/wikipedia-able or culture specific ([url=http://en.wikipedia.org/wiki/Damocles]SWORD : DAMOCLES[/url]:: BED : ?).

You can find the super-test
[url=http://www.eskimo.com/~miyaguch/power.html]Here[/url].

EDIT: The main page is [url=http://www.eskimo.com/~miyaguch/hoeflin.html]here[/url].


complex numbers
combinatorics proposed
combinatorics

use the following:
  $(1+x)^n=\sum_{i=0}^n \binom{n}{k} x^k$;plug in $x=1,\omega,\omega^2$, where $\omega^3=1$( primitive 3rd root of unity). if $A_i=\sum_{k\equiv i(\mod 3)} \binom{n}{k}$ then we get 3 linear equations frm the above in the variables $A_0,A_1,A_2$.what you want is $A_1$. solve the equations to get what you want.

My teacher's solution also uses complex numbers . It's similar to yours . But I think a simple problem will have an elementary proof ;)


algebra polynomial trigonometry inequalities

Let $a_{0}, a_{1}, \cdots ,  a_{n} \in \mathbb{C}$ and $f: \mathbb{C} \to \mathbb{C}$, $f(z)=a_{0}+a_{1} \cdot z +a_{2} \cdot z^2+ \ldots +a_{n} \cdot z^n$. It is known that $|z|=1$ implies $|f(z)| \leq 1$. Prove that $|a_{k}| \leq 1$, $\forall k \in \{ 0,1, \ldots, n\} $.

Put $z=cis \theta$, then we have
$(\sum_{i=0}^n a_i \cos i \theta)^2 + (\sum_{i=0}^n a_i \sin i \theta)^2 \le 1$.
For all $k \in (0,1,...,n)$, we have
$a_i^2 = a_i^2 cos^2 i \theta + a_i^2 sin^2 i \theta \le (\sum_{i=0}^n a_i \cos i \theta)^2 + (\sum_{i=0}^n a_i \sin i \theta)^2 \le 1$.

[quote="Soarer"]$a_i^2 cos^2 i \theta + a_i^2 sin^2 i \theta \le (\sum_{i=0}^n a_i \cos i \theta)^2 + (\sum_{i=0}^n a_i \sin i \theta)^2$.[/quote]

Why is this inequality true? Am I missing trivial things  :?

See [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=elegant&t=1539]here (problem 3)[/url] for a solution to a stronger statement.


If there is anyone who needs/wants a spot to take the AMC 12A or B, I had to order extra tests to take it so it will be given in the early afternoon of the day of each test (Tuesday, February 9th for the A and Wednesday, February 24th for the B) in West Chester (near Cincinnati). Also, there will be a $ \$$10 fee to pay for the test. If you need more information (exact location, etc.) PM me. Either reply here or PM me if interested. Thanks.

Wait...what?
You have the real thing? Um..isn't that kind of cheating though? knowing the problems ahead of time..?

No, I'm going to have extra tests when I take it. They come in 10-packs.

ah..i see.
ok sorry, but I don't want them

first, ,I'm too young.lol


$ a,b,c,d \geq 0$ and $ a^2\plus{}b^2\plus{}c^2\plus{}d^2\equal{}1$ prove:

$ a^3\plus{}b^3\plus{}c^3\plus{}d^3\plus{}abc\plus{}abd\plus{}acd\plus{}bcd\leq 1$

[quote="zaya_yc"]$ a,b,c,d \geq 0$ and $ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \equal{} 1$ prove:

$ a^3 \plus{} b^3 \plus{} c^3 \plus{} d^3 \plus{} abc \plus{} abd \plus{} acd \plus{} bcd\leq 1$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=99927
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=90039


A stone falls from a ledge and takes 8 seconds to hit the ground.
The stone has an original velocity of 0 m/s. How tall is the ledge?

A. 1,254.4 meters

B.313.6 meters

C.78.4 meters

D.39.2 meters

This is physics.

[hide="answer"]

Gravitational acceleration is approx. 9.8 m/ss

$distance = velocity_{initial} \times time + \frac{1}{2} \times acceleration \times time^2$

Currently known articles:

acceleration: 9.8
time: 8
Initial velocity: 0

Plugging them in, I get,

$d = 0 \times 8 + \frac{1}{2} \times 9.8 \times 8^2$

$d = \frac{1}{2} \times 9.8 \times 64$

$d = \frac{1}{2} \times 627.2$

$d = 313.6$

The height of the cliff is 313.6 meters.

Answer: B

[/hide]

[hide]Since $d=\frac{1}{2}at^2$ when the initial velocity is 0 (and a=9.8), so we have

$d=\frac{1}{2}(9.8)(8^2)=313.6m$. [/hide]

And hwenterprise, you forgot to square the 8  ;)


Given positive real x, y, z, x + y + z = 1. Prove:

(x^(-2)-x)(y^(-2)-y)(z^(-2)-z) larger than or equal to (26/3)^3

In my opinion, it is not hard to prove  :wink:
The inequality is equivalent to:
\[ (1 \minus{} x^3)(1 \minus{} y^3)(1 \minus{} z^3)\ge \left(\frac {26}{3}\right)^3x^2y^2z^2
\]

\[ \iff 1 \plus{} (x^3y^3 \plus{} y^3z^3 \plus{} z^3x^3) \minus{} (x^3 \plus{} y^3 \plus{} z^3) \minus{} x^3y^3z^3\ge \left(\frac {26}{3}\right)^3x^2y^2z^2
\]
Using the following result. Note that $ x \plus{} y \plus{} z \equal{} p,xy \plus{} yz \plus{} xz \equal{} q,r \equal{} abc$:
\[ x^3y^3 \plus{} y^3z^3 \plus{} z^3x^3 \equal{} 1 \minus{} 3q \plus{} 3r
\]

\[ x^3 \plus{} y^3 \plus{} z^3 \equal{} 1 \minus{} 3qr \plus{} 3r^2
\]
We will prove an inequality with two variables is $ q,r$.
\[ q^3 \plus{} 3r^2 \minus{} 3r \minus{} r^3 \plus{} 3q(1 \minus{} r)\ge \left(\frac {26}{3}\right)^3r^2
\]
On the other hand, using AM-GM and Schur's inequality we have: $ r\le \frac {1}{27}$ and $ q\ge 9r$. So we work only with $ 1$ variable is $ r$. Remain part what is easy to work. The inequality is equivalent to:
\[ \frac {8r(27r \minus{} 1)(91r \minus{} 81)}{27}\ge 0
\]
It is always true for all $ r\le \frac {1}{27}$. The inequality was proved.

I found a stronger one  :D
Let $ x,y,z\ge 0$ such that $ x\plus{}y\plus{}z\equal{}1$. Prove that:
\[ (1 \minus{} x^3)(1 \minus{} y^3)(1 \minus{} z^3)\ge \left(\frac {26}{3}\right)^3\frac{(x\plus{}y\plus{}z)^6}{3^6}
\]

I hope that the following inequality is also true... I will check later because I am going to school. If it is true, I think it is also easy  :)
Let $ x,y,z\ge 0$ such that $ x \plus{} y \plus{} z \equal{} 1$. Prove that:
\[ (1 \minus{} x^3)(1 \minus{} y^3)(1 \minus{} z^3)\ge \left(\frac {26}{3}\right)^3\frac {(x^2 \plus{} y^2 \plus{} z^2)^3}{27}
\]
Can you disprove above inequality?  :D

[quote="zaizai-hoang"]I hope that the following inequality is also true... I will check later because I am going to school. If it is true, I think it is also easy  :)
Let $ x,y,z\ge 0$ such that $ x \plus{} y \plus{} z \equal{} 1$. Prove that:
\[ (1 \minus{} x^3)(1 \minus{} y^3)(1 \minus{} z^3)\ge \left(\frac {26}{3}\right)^3\frac {(x^2 \plus{} y^2 \plus{} z^2)^3}{27}
\]
[size=100][color=darkblue]Can[/color][/size] [size=150][color=green]you[/color][/size] [size=200][color=red]disprove[/color][/size] [size=75][color=violet]above[/color][/size] [size=59][color=orange]inequality[/color][/size] ?  :D[/quote]

[size=150]Yes , I can[/size]  :lol:

[quote="zaizai-hoang"]I found a stronger one  :D
Let $ x,y,z\ge 0$ such that $ x \plus{} y \plus{} z \equal{} 1$. Prove that:
\[ (1 \minus{} x^3)(1 \minus{} y^3)(1 \minus{} z^3)\ge \left(\frac {26}{3}\right)^3\frac {(x \plus{} y \plus{} z)^6}{3^6}
\]

[size=100][color=darkblue]Can[/color][/size] [size=150][color=green]you[/color][/size] [size=200][color=red]disprove[/color][/size] [size=75][color=violet]above[/color][/size] [size=59][color=orange]inequality[/color][/size] ?

[/quote]

[size=150]Sure[/size]  :lol:

Yes, I am wrong. I hurried  :blush: My mistake is:
\[ (27r\minus{}1)(66339r^2\plus{}270r\plus{}2197)\ge 0\]
It is wrong  :D Have any result from above inequality?

[quote="zaizai-hoang"]Yes, I am wrong. I hurried  :blush: My mistake is:
\[ (27r \minus{} 1)(66339r^2 \plus{} 270r \plus{} 2197)\ge 0
\]
It is wrong  :D Have any result from above inequality?[/quote]
Replace $ x\plus{}y\plus{}z$ by this one $ xy\plus{}yz\plus{}zx$ the inequality is true  :wink: , Hoang !


induction
number theory
number theory solved

Show that any positive integer has at least as many positive divisors of the form zk + 1 as of the form zk - 1, z \in {3,4,6,8,12,24}. Show also that the proposition holds for no other pairs of arithmetic progressions.

[i]Mikls Schweitzer competition 1953 problem 5[/i]

Sorry but do u mean the if n is a positive integer then
   card {d |n :  d==1(modz) }=card{d|n : d==-1 (modz) }
for all z in {3,4,6,8,12,24} ?
In the case i see the proof for the case z=4 in the book "250 problems in Elementary number theory " of Waclaw Sprienski and it is the last problem of the Divisbility section as i remember!It prove by induction so you can try by this method!
tHE CONVESE ,i have no...comment.It needs time to think!

Hi,

I see you are well informed about VietNam NMO. There are also some mathlinks users who are joining it.  :D

At first I do not have the book you refer to. Second I do not know what you mean when you refer to card: Is it kind of set for the elements of a certain relation or something like that ?

In order to clarify matters. Each time you consider a postive integer x, your z will be fixed (constant) and k runs through all integers generating all the positive divisors zk + 1 and zk - 1 of the positive integer x. And your task is to show that this theorem is true if and only if z \in {3,4,6,8,12,24}.

i think he means card as in cardinality, or number of elements of the set.


Protons are rotating in circle with radius R with initial velocity $v_{0}$.Initial current in the system is $I_{0}$.The system is put into the magnetic field where the magnetic flux is changing with constant speed like this:

$k=\frac{\delta \phi}{\delta t}$

k is constant

Find the current in the system and the velocity of protons after N number of turns.

Each proton will increase its kinetic energy by $keN$, so the new speed will be
\[v = \sqrt{v_{0}^{2}+\frac{2ekN}{m}}.\]
Since $I = cev$, where $c$ is the concentration fo the protons, we get
\[I = I_{0}\sqrt{1+\frac{2ekN}{mv_{0}^{2}}}.\]


Given a triangle ABC such that AB=13, BC=14, and AC=15, draw a cevian from vertex A to a point D on side BC. Let the incircles of triangles ADB and ADC be circles O and P respectively. If the two circles are tangent at a point X on the cevian AD, what is the sum of the distances from point X to the sides of AB and AC?
[hide]
i had a solution though it was very inelegant and i had taken the cevian to be an altitude to simplify things because obviously the result is independent of the cevian chosen. However this involved a lot of nasty calculations. Can anyone find the answer for any general cevian(not a special cevian like altitude or angle bisector)[/hide]

I am not able to get the sum.
@ befuddlers can you tell me how you got the solution?
I too tried by taking the cevian to be the altitude, but its too lengthy.

i dont think it will work for any cevian.
if we take point D about 1cm away from C (to the left, i.e. 13cm away from B) , the resulting 2 circles will be tangent at 2 different points to the line AD. so i think the question first asks you to identify the [i]particular[/i] cevian where this property will hold.

this can be easily done with a little bit of reasoning, let us first take an isosceles triangle ABC with AB = AC, then join A to D - the midpoint of EG - now, we know that in this case the circles WILL be tangent to each other and the line at a single point, this line is the altitude as well as the median of the aforesaid triangle. now we only have to find out which of the two is the one cevian which will always work for non-obtuse scalene triangles (as that is the triangle given to us). so it is either the median or the altitude. the median does not work (i checked by construction :rotfl: )
hence it is the altitude. actually this might not help, but at least it assures you that you were on the right track

[quote="accuracy"]
this can be easily done with a little bit of reasoning, let us first take an isosceles triangle ABC with AB = AC, then join A to D - the midpoint of EG - [/quote]

that was "midpoint of BC" and   :oops:


Prove that for all $n \in N$ there exist $a,b,c$satisfy $a|bc$and $n^{2}<a,b,c<n^{2}++n+3\sqrt{n}$

[quote="ZetaX"]$a,b,c$ distinct¿[/quote]
Oh yeah
sorry i'm mistake


An integer $ x$, with $ 10 \le x \le 99$, is to be chosen. If all choices are equally likely, what is the probability that at least one digit of $ x$ is a $ 7$?

$ \textbf{(A)}\ \frac19\qquad
\textbf{(B)}\ \frac15\qquad
\textbf{(C)}\ \frac{19}{90}\qquad
\textbf{(D)}\ \frac29\qquad
\textbf{(E)}\ \frac13$

2 digit numbers with 7 are: 17,27,...,97 and 70,71,...,79 but 77 occurs twice so there are 18 those numbers, giving us the probability $ \frac{18}{90}\equal{}\frac 15$
B

There are $ 99\minus{}10\plus{}1\equal{}90$ total possibilities.  In order for at least one of the digits to have a $ 7$ in it, it must be in the units or tens digit.  Let's look at where 7 is in the tens digit.  There are $ 79\minus{}70\plus{}1\equal{}9\plus{}1\equal{}10$ possibilities here.  As for the units digit, there are $ 8$ possibilities: $ 17, 27, 37, 47, 57, 67, 87, 97$ (notice how we don't count $ 77$; we've already counted this when we counted all of the numbers with $ 7$ as the tens digit).  Hence, the probability that there is a $ 7$ in the number is $ \frac{18}{90}\equal{} \frac{1}{5}$ $ \boxed {D}$


Let $ABCD$ be a rhombus with angle $A = 120$ and $P$ is a point in its plane. Prove that $PA+PC>\frac{BD}{2}$

[hide="simple way"]
You know that the shortest possible length of PA + PC is when P is on AC. So, all you have to prove is that AC > BD/2. That's not that hard. Set the length of one side of the rhombus to x. Since A is 120 degrees, it's just split into two equilateral triangles.

AC = x
BD = x :sqrt: 3.

Assuming x is positive (which it must be)...

1 > ( :sqrt: 3 )/ 2
x > (x :sqrt: 3)/2
AC > BD/2

And we all know that AC is the lowest possible value of PA + PC, so if AC is greater than BD/2, PA + PC must be.

PA + PC > BD/2
[/hide]


function
LaTeX
calculus
derivative
real analysis
real analysis unsolved

For each integer \[n \ge 1 \]
let \[{\rm f}_{\rm n}\]: R -> R be the monotone increasing function

\[{\rm f}_{\rm n}(x) ={\rm }\left\{ \begin{array}{l}\frac{{-1}}{{n(n+1)}}{\rm ; where x < }-{\rm }\frac{{\rm 1}}{{\rm n}}\\{\rm 0; where }-\frac{1}{n}\le x < \frac{{\rm 1}}{{\rm n}}\\ \frac{1}{{n(n+1)}}{\rm ; where }\frac{{\rm 1}}{{\rm n}}\le{\rm x }\\ \end{array}\right. \]

Prove that the sum F = \[\sum_{n = 1}^\infty{f_{n}}\]

is again a monotone increasing function, and is differentiable at all points of
\[{\rm R\backslash (\{ }{\textstyle{{\rm 1}\over{\rm n}}}{\rm \}}_{{\rm n}\ge{\rm 1}}\cup \{{\textstyle{{-1}\over n}}\}_{n \ge 1}) \]

with F ' (0) = 1
\[{\rm eventhough f '}_{\rm n}(0) = 0{\rm for all n}\ge{\rm 1}\]

$\LaTeX$ cuts out spaces if you're not in a text environment; you should adjust your code to force spaces where appropriate, i.e "$\text{where }x<\frac{-1}n$"

Just calculate the sum explicitly: for $x>0$, $f(x)=\sum_{n\ge\frac1x}\frac1{n(n+1)}=\sum_{n=m}^{\infty}\frac1n-\frac1{n+1}=\frac1m$, where $m$ is the least integer greater than or equal to $\frac1x$. (negative $x$ are similar) This is obviously differentiable with derivative zero between the jump discontinuities. At the origin, apply the definition of the derivative.

I don't see Fubini having much to do with this.

It was explained by my professor that this problem was kind of a warning for what we call Fubini's theorem, I didn't really understand the comment either, so omit the remark that it relates to Fubini's theorem.

Here's the link: the general "Fubini" principle is that we can exchange limit operations (in this case, a sum and a derivative) if the quantities involved are nice enough. The caveat is that "nice enough" does matter, as examples like this show.

Fubini's theorem could be used to prove that this interchange works under certain conditions on the $f_{n}$. This example doesn't come close to meeting those conditions.


Euler
geometry
circumcircle
geometric transformation
reflection
conics
incenter

Let $ ABC$ be a given triangle and let $ \rho_{A}$, $ \rho_{B}$, $ \rho_{C}$ be the lines through the vertices $ A$, $ B$, $ C$ and parallel to the Euler line $ OH$, where $ O$ and $ H$ are the circumcenter and orthocenter of $ ABC$. Let $ X$ be the intersection of $ \rho_{A}$ with the sideline $ BC$. The points $ Y$, $ Z$ are defined analogously. If $ I_{a}$, $ I_{b}$, $ I_{c}$ are the corresponding excenters of triangle $ ABC$, then the lines $ XI_{a}$, $ YI_{b}$, $ ZI_{c}$ are concurrent on the circumcircle of triangle $ I_{a}I_{b}I_{c}$.

Do you really expect the concurrency to vanish if the parallel lines thru vertices
   have a direction parallel to OI line or in fact any other real (or imaginable?)  line ?

   M.T.

Dear armpist,

The concurrence obviously follows from the Cevian Nest Theorem. However, this was not the purpose of the problem. The interesting thing (at least for me) is the position of the perspector. Note also that this perspector lies on the line $ E'H$, where $ E'$ is the Euler reflection point of triangle $ I_{a}I_{b}I_{c}$.

Regards,
Cosmin

[b]Generalization[/b] [b]:[/b] [color=#00f]Given a $ \triangle ABC $ with points $ P, Q. $ Let $ \triangle P^aP^bP^c $ be the anticevian triangle of $ P $ WRT $ \triangle ABC $ and let $ T $ be a point lying on the tripolar of $ Q $ WRT $ \triangle ABC. $ Then the Cevian quotient $ (T/P) $ lies on the circumconic $ \mathcal{K}_{(Q/P)} $ of $ \triangle P^aP^bP^c $ with perspector $ (Q/P) $ WRT $ \triangle P^aP^bP^c. $[/color]

[b]Proof[/b] [b]:[/b] Let $ \triangle Q_aQ_bQ_c, \triangle T_aT_bT_c $ be the cevian triangle of $ Q, T $ WRT $ \triangle ABC, $ respectively and let $ V_b, V_c $ be the intersection of $ CA, $ $ AB $ with the tripolar of $ T $ WRT $ \triangle ABC. $ First, it's obvious that $ BV_b, CV_c $ passes through the harmonic conjugate $ X $ of $ T $ WRT $ A, T_a. $ Furthermore, $ X $ lies on $ Q_bQ_c $ since $ T $ lies on the tripolar of $ Q $ WRT $ \triangle ABC, $ so $$ P^b (A, Q_b; C, V_b) = P^c (A, Q_c; V_c, B) \Longrightarrow (Q/P) \in \varsigma $$ where $ \varsigma $ is the tripolar of $ (T/P) $ WRT $ \triangle P^aP^bP^c, $ hence we conclude that $ (T/P) \in \mathcal{K}_{(Q/P)} $ $ \color{blue} ^{[1]} $ $. $

[color=#00f][1][/color] [url=https://artofproblemsolving.com/community/c284651h1687256]Blog - Connection between Inconic and Circumconic[/url] ([b]Lemma 1[/b])


:censored: or :cursing:
I'd say :cursing:

[quote="1=2"]gotta love censored.[/quote]

censored is annoying.

censored is boring. cursing is animated  :D

Yep.  I agree.  Censored is a bit bland.

WAT DA :censored: YOU :cursing: :censored: :cursing: !!!!!!!!

... i am not part of this conversation.

i was not here at all


How  to   include   picture  in  pdfscreen?
For  example,  I   haved  read    the  Manual, and   there  is  a  picture  on the   right  top  corner    of  the  pdf(on  the upper part of  the  panel).
I  don't konw  how  to get  the  output.

An  another   example,  the  picture  on  the  file  widepanel.pdf.

what    is  the  code ?

Many  thanks

It tells you how to do so in the manual on page 9! [code]\includegraphics[width=.75in]{univ}\par\vfill[/code]


[hide]1000 prime factorized equals $2^{6}\cdot5^{6}$.  Now to calculate the factors we add one to each of the exponents and multiply.  So $7\cdot7=\boxed{49}$.  [/hide]

[quote="i_like_pie"][hide]The number is 100000, not 1000000. $100,000 = 10^{5}= 2^{5}\cdot 5^{5}$. The correct answer is $36$.[/hide]

[size=75][color=darkred]Try to avoid putting text into LaTeX whenever possible. Thanks.
-nebula42[/color][/size][/quote]
oops!  :oops:

[hide]36 divisors[/hide]

[size=75][color=darkred]Don't forget to hide your answers. Thanks.
-nebula42[/color][/size]

Not so much a proof, but if you think about it, all factors of $a^{b}c^{d}e^{f}...$ can be made from a's c's and e's. So for example $2^{5}(5^{5})$ 2 and 5 make all of the factors. So your choices are $2^{0}2^{1}2^{2}2^{3}2^{4}2^{5}$ for 2 which gives you 6 choices. This same thiinking can be used for 5. All factors are of 100000 are created by multiplying one of the 2s and one of the 5s. So you have 6 choices for each. 6(6)=36

[quote="Arvind_sn"]Yes, that's what I got as well. :)  However, can anyone find a way to prove that for any number $a^{b}*c^{d}*....$ where a and c (and other prime factors) are prime, The number of divisors is b+1 times C+1 times...(so on, so on)? I have been working with that rule for a year now and I still do not have the proof.[/quote]
To generalize what bpms wrote, let's say $n={p_{1}}^{e_{1}}{p_{2}}^{e_{2}}{p_{3}}^{e_{3}}\dots$ (where the $p_{i}$ are primes). Consider the number $x ={p_{1}}^{d_{1}}{p_{2}}^{d_{2}}{p_{3}}^{d_{3}}\dots$, where $d_{i}= 0, 1, 2, \dots,$ or $e_{i}$ for any $i$. Then we can see that any number of this form is a factor of $n$, and any factor of $n$ can be expressed in this form. This means that the number of factors of $n$ is equal to the number of possibilities for $x$, which is $(e_{1}+1)(e_{2}+1)(e_{3}+1)\dots$.

For a specific example of this, see bpms's post above.

[hide="solution"]factor it:

2^5*5^5

there are 36 divisors[/hide]


Suppose that $\tau_{1}, \tau_{2}$ are two positive measures on $R$ with its Borel sets such that for all open intervals $I$ of $R$ of finite length, we have $\tau_{1}(I)\leq\tau_{2}(I)<\infty$. Then $\tau_{1}(X)\leq\tau_{2}(X)$ for all Borel sets $X$.

This is because such measures are outer regular because of the finiteness assumption.


Given $ a,b,c>0$ and $ a\plus{}b\plus{}c \equal{}1$.Prove that:
$ \frac{b}{\sqrt{a^2\plus{}2b}}\plus{}\frac{c}{\sqrt{b^2\plus{}2c}}\plus{}\frac{a}{\sqrt{c^2\plus{}2a}} \le \frac{3}{\sqrt{7}}$ :)

We only need to prove:
$ \sum\ \frac{a^2}{a^2\plus{}2b} \geq\ \frac{3}{7}$ (obviously trues again)   :)

[quote="nguoivn"]We only need to prove:
$ \sum\ \frac {a^2}{a^2 \plus{} 2b} \geq\ \frac {3}{7}$ (obviously trues again)   :)[/quote]
Yes :lol:
@nguoivn[hide] anh biet het nhung dang toan nay roi a :) [/hide]

@quykhtn-qa1[hide]anh và anh nhocnhoc hình như thích kiểu dòng chữ ẩn quá,anh quy ơi anh tên chính xác là gì?Học lớp nào ạ?khối nào ạ[/hide]


inequalities
analytic geometry
real analysis
real analysis unsolved

Suppose $ 1<p<\infty$. Show that $ f_n\to f$ weakly in $ \ell^p$ (that is, $ \sum_{k=1}^\infty f_n(k)g(k)\to \sum_{k=1}^\infty f(k)g(k)$ for all $ g\in \ell^q$) iff $ f_n\to f$ pointwise and $ \sup_n\|f_n\|_p <\infty$.

I have some things worked out, and ideas about others, but almost nothing is actually coming together.

For example, if $ f_n$ does not converge pointwise to $ f$, then there is at least one point $ k_i$ such that $ |f_n(k_i)-f(k_i)|$ doesn't get small. Thus you can just choose $ g(k)=\begin{cases} 1 \ k=k_i \\ 0 \ else \end{cases}$ which is definitely $ q$-power summable since it will equal 1. We don't get the sums converging then since $ \big|\sum_{k=1}^\infty f_n(k)g(k) - \sum_{k=1}^\infty f(k)g(k)\big| = |f_n(k_i)-f(k_i)|$.

The main thing I think is that I don't know what to do with that sup. Something like that flashes signs of Banach-Steinhaus. Also, when bounded things summable things of conjugate exponents, there should be a nice Holder inequality in there. It has failed to produce useful inequalities when I try it though.

[quote="HilbertThm90"]Something like that flashes signs of Banach-Steinhaus.[/quote]
Which says that a pointwise bounded collection of operators (on the right kind of spaces, which we have) is uniformly bounded. So consider $ \{f_n\}$ to be a collection of linear operators acting on $ \ell^q.$ Since $ \langle f_n,g\rangle$ converges for each "point" $ g,$ that's your pointwise boundedness. You can finish it from there.

There's also the converse to consider: suppose $ \|f_n\|_p\le M$ for some $ M$ and $ f_n(k)\to f(k)$ for all $ k.$ Can you prove that $ \sum f_n(k)g(k)\to\sum f(k)g(k)?$

For the direct implication:
you proved that the pointwise convergence follows and you're left with that uniform boundedness. You apply Banach-Steinhaus to the maps $ f_n\in (l^q)^*\cong l^p$ ($ f_n(g)$ are bounded for every $ g\in l^q$) and you get the uniform bound.
For the converse: take an arbitrary $ g\in l^q$. Then look at its "coordinates" and find an $ N$ such that from that point on the sum of the $ q$th powers is less than $ \varepsilon$ (i.e. $ \sum_{k\geN + 1} |g(k)|^q\le \varepsilon$). Denote by $ T$ the maximum of the $ g(k),k = 1,...,N$
Now since $ f_n$ converge pointwise to $ f$ choose an $ n_0$ big enough such that $ |(f_n - f)(k)| < \delta,\, \forall n\ge n_0, k\in\{1,...,N\}$.
Now $ |(f_n - f)(g)| \le |\sum_{k = 1}^N(f_n - f)(k)g(k)| + |\sum_{k\ge N + 1}(f_n - f)(k)g(k)|\le$
 $ \le T\delta + \varepsilon\sup\parallel{}f_n - f\parallel{}_p$ and here you use the fact that the sup is finite and you're done (I also used [url=http://en.wikipedia.org/wiki/Holder_inequality]Holder's Inequality[/url] for)
[edit] I haven't seen your post Kent...

So I had most of those things scratch all over my page, but for some reason I just didn't tie it together properly. I think I should have converted to better notation to think in terms of linear operators instead of keeping summations everywhere.  :maybe:

So the direct implication relies very heavily on $ (\ell^q)^*\cong \ell^p$, thus for $ p\equal{}1$ the same argument doesn't work (not that that case was part of the problem).

For $ p\equal{}1,$ the forward argument (the one using Banach-Steinhaus) still works. It is still true that $ \ell^1$ is isometrically contained in $ (\ell^{\infty})^*,$ and that's enough. It's the converse argument (Kuba's argument) that fails.

Right away, you see that Kuba used the existence of an $ N$ such that $ \sum_{k>N}|g(k)|^p<\epsilon.$ You know we can't use that exact idea.

Consider $ f_n$ such that $ f_n(k)\equal{}\begin{cases}\frac1n,&1\le k\le n\\0,&\text{otherwise}\end{cases}.$

$ \|f_n\|_1\le 1$ so the norms are bounded, and $ f_n$ tends to zero pointwise. But $ f_n$ does not tend to zero weakly, as can be seen by testing it against a $ g$ which is identically $ 1.$


Let $ a,b,c > 0$ and $ x\in\Bbb R$ prove that
\[ \frac {b \plus{} c \minus{} a}{b \minus{} a}\cdot\frac {b^x \minus{} a^x}{(b \plus{} c)^x \minus{} a^x} > \frac {b^{x \minus{} 1}}{(b \plus{} c)^{x \minus{} 1}}
\]
:D

Sorry mine is $ x\in \Bbb R$ but yours needs $ r>0$

There might be some conditions missing :wink:
If $ a\equal{}b$ or $ b\plus{}c\equal{}a$ or $ x\equal{}0$ for example, the problem doesn't make any sense.
Daniel


geometry
incenter
search
geometry unsolved

Show that Incenter-Gergonne point line and 3 similar ex-center - ex-Gergonne point
   lines are concurrent at ? point.

   M.T.

   Happy Holidays to all.

Haha maybe it was posted http://www.mathlinks.ro/viewtopic.php?search_id=1104335482&t=214545 but no one is interested in !

Sorry for barging onto your turf.  Hopefully this time a solution will be found.

    M.T.

Dear Armpist and Mathlinkers,
this is a nice problem.
The incenter-Gergonne line is the Soddy's line and passes through the de Longchamps's point noted L. I am sure that Armpist knows this result.
The three extraversion comme from Hermes, a german geometer, and pass also through L.
The synthetic proof is not so easy... if we want to explain all the details. An article concerning this result will appear next year on my website.
Sincerely
Jean-Louis


induction
linear algebra
matrix
ARML
geometry
rectangle
function

You COULD brute-force it using Binet's formula.

Without being immodest, I love this proof.

Use the matrices for the Fibonacci sequence, that is

$\left( \begin{array}{ccc}
F_{n+1} & F_n \\
F_n & F_{n-1} \end{array} \right)=\left( \begin{array}{ccc}
1 & 1 \\
1 & 0 \end{array} \right)^n$

Now, look at this!

$\left( \begin{array}{ccc}
F_{2n+1} & F_{2n} \\
F_{2n} & F_{2n-1} \end{array} \right)=\left( \begin{array}{ccc}
1 & 1 \\
1 & 0 \end{array} \right)^n\left( \begin{array}{ccc}
1 & 1 \\
1 & 0 \end{array} \right)^n=\left( \begin{array}{ccc}
F_{n+1} & F_n \\
F_n & F_{n-1} \end{array} \right)^2=\left( \begin{array}{ccc}
F_{n+1}^2+F_n^2 & \times \\
\times & \times \end{array} \right)$

and we see that $F_{2n+1}=F^2_{n+1}+F^2_n$

I did practically the same thing but I reduced the given expression to another one before I tried using the matrices. The method you gave is brilliant! :lol:

oo nice solution, im not very familiar w/ the fibonacci matrices tho

i proved it using some induction, which was much more messy soooo i wont bother typing it out unless someone really wants to see it

im actually not even sure wat i did counts as induction but w/e haha  :P

Well I'm not sure you can do it unless you use a double induction, that is claiming two statements both work at the same time and then using induction on both of them after which you use the first to prove the second and vice versa. Could you type out your method? I originally reduced the expression to $f_n(f_{n-1}+f_{n+1})=f_{2n}$ which I then saw worked with the matrices by using the same expansion that blahblah gave except looking at the lower left corner.

[quote="DPopov"]We define the Fibonacci sequence to be $f_0=0$, $f_1=1$ and for every $n$, $f_n=f_{n-1}+f_{n-2}$.

Prove $f_{n+1}^2+f_{n}^2=f_{2n+1}$[/quote]

Fibonacci numbers count sets of positive integers with largest element equal to $n$ and no two elements are neighbors.
If the largest element is $2n+1$ there are two cases, according to whether the middle element $n+1$ is in the set or not.

Prove, by very straightforward induction, $\displaystyle f_{n+m} = f_{n-1}f_m+f_nf_{m+1}$

then substitute $m=n$, simplify the expression, and tah dah!

[quote="pkothari13"] Prove, by very straightforward induction, $f_{n+m} = f_{n-1}f_m+f_nf_{m+1}$ [/quote]

The combinatorial argument for this is the same as above:
Of the Fibonacci sets with largest element $m+n$, some contain $m$ and some do not.

The algebraic argument is to look at one of the matrix entries of ${F^m}{F^n} = F^{m+n}$.

I didn't see the straightforward induction as being easy, maybe I made a careless mistake somewhere. ;)

$f_{n+m} = f_{n-1}f_m+f_nf_{m+1}$

if it's true for $m$, then its true for $m+1$ and $m+2$

$f_{n+m+1} = f_{n-1}f_{m+1}+f_nf_{m+2}$

add these two equations lining up each term and by using the definition of fibonacci series, you'll get

$f_{n+m+2} = f_{n-1}f_{m+2}+f_nf_{m+3}$

so now you know that this formula is true, substitute $m=n$ and you'll get

$f_{2n} = f_{n-1}f_n + f_nf_{n+1}$

$f_{2n} = f_n(f_{n-1}+ f_{n+1})$

because $f_n = f_{n+1}-f_{n-1},$ you get

$f_{2n} = (f_{n+1}-f_{n-1},)(f_{n-1}+ f_{n+1})$

and then

$f_{2n} = f_{n+1}^2 - f_{n-1}^2$

(EDIT) AHHHHH!! shoot! i proved the wrong thing!!!!!!!!  :censored:  :cursing:

There's an interesting way to prove this using combinatorics.

Generally, $f_{n+1}$ counts the number of tilings of a $n \times 1$ strip of squares using squares or dominos or both. (You can prove it in many different ways.)

Consider counting the number of $2n$-tilings. There are two different ways to count it.
1. As mentioned  earlier, the number of $2n$-tilings is $f_{2n+1}$.
2. Consider two cases: whether there is a domino right in the middle of the tiling (so that the tiling does not break in the middle). There are $(f_n)(f_n)=f_n^2$ cases that the tiling does not break in the middle. On the other hand, there are $(f_{n+1})(f_{n+1})=f_{n+1}^2$ cases that the tiling does break in the middle. So the total number of tilings is $f_n^2+f_{n+1}^2$.

Therefore, we have $f_n^2+f_{n+1}^2=f_{2n+1}$.
(Source: 2004 Chicago ARML team practice)

Sorry, it's not very easy to explain such a proof, so correct me if there is any mistake or anything unclear.

frt, that proof was [i]awesome[/i].

You know any methods to prove that the number of tilings of an n X 1 rectangle is f_n+1 other than using a recursion?

[quote="WarpedKlown1335"]frt, that proof was [i]awesome[/i].[/quote]Thanks :)

[quote="WarpedKlown1335"]You know any methods to prove that the number of tilings of an n X 1 rectangle is f_n+1 other than using a recursion?[/quote]I've heard that there are many ways to prove it, but I'm afraid I can prove it only using a recursion :(

I've seen that proof in Zeitz!!! I agree, it's an awesome proof  :lol:

Interestingly, there's another relationship between $f_n$ and tilings. (Maybe some people already know this one :) )

$f_{n+1}$ is a number of ways to tile a $2\times n$ checkerboard with $2\times 1$ dominoes. For example, $f_4=3$ is a number of ways to tile a $2\times 3$ checkerboard.

Reference: [url=http://mathworld.wolfram.com/Domino.html]http://mathworld.wolfram.com/Domino.html[/url]

Yeah, you can either have a vertical $1$x$2$ tile at the end or two horizontal $1$x$2$ tiles which gives you the fibonacci recursion.

You might like to experiment with this method which expresses membersof the Fibonacci Sequence in terms of hyperbolic fuctions.

Write( !+Sqrt5)/2 =ExpT, then  -(1-Sqrt5)/2 = Exp(-T)

So we have sinh(T) = 1/2 and cosh(T)=(Sqrt5)/2

And for odd n, F(n) =2/Sqrt5.cosh(nT) and for even n, F(n) =2/Sqrt5.sinh(nT), where F(n) is the n-th Fibonacci number.
Now apply properties of the hyperbolic functions -addition formulae, double angle formulae,in fact all the hyberbolic equivalents of the standard trig formulae.The only drawback is that two separate calculations are necessary -one for n odd the other for n even.

Source Mathematical Gazette 2001, page 477

This is pretty much Binet's Formula right?