Find all polynomials $P(x)$ with real coefficients such that there exists a polynomial $Q(x)$ with real coefficients that satisfy
\[P(x^2)=Q(P(x)).\]

Help me! I know this problem is really difficult, but there is a person who can anwser this question for me?

If $P$ is constant, that is, $P(x)=c$ for some $c$, then we can take $Q(x)=x$ and we have $Q(P(x))=P(x)=c=P(x^2)$. So constant polynomials work.

Now let $n>0, m$ be the degrees of $P, Q$ respectively. The LHS has degree $2n$, while the RHS has degree $mn$. Then $m=2$. So $Q(x)=a(x+b)^2+c$ for some constant reals $a, b, c$. Now, $P((-x)^2)=P(x^2)$, so $Q(P(-x))=Q(P(x))$ for all $x$. Then $a(P(x)+b)^2+c=a(P(-x)+b)^2+c$. It follows that $P(x)+b= \pm (P(-x)+b)$, for each $x$. If $P(x)+b=P(-x)+b$ for infinitely many $x$, then the polynomials $P(x), P(-x)$ are identical, and $P(x)=R(x^2)$ for some $R$. Otherwise $P(x)+P(-x)+2b=0$ for infinitely many $x$, so $P$ is almost odd.

Now note that, if $P(x)=R(x^2)$ is a good polynomial, then $R(x)$ must be too (just let $x^2=y$ and note that the coefficients of the terms on both sides must keep being equal). So, if $P(x)$ is even, take $P_2(x)=R(x)$. If it is even too, take $P_3(x)$ and so on, until $P_k(x)=S(x)$ is not even. Then, remember: $S(x)+S(-x)+2b=0$. Then $S(x)+b, S(-x)+b$ add up to $0$; so $S(x)+b$ is odd, that is, $S(x)+b=xT(x^2)$, $S(x)=xT(x^2)-b$.

Finally, this yields $ax^2(T(x^2))^2+c=x^2T(x^4)-b$. So $c=-b$ and $a(T(x^2))^2=T(x^4)$. Let $\omega$ be a complex root of the LHS. Then it's obviously a root of the RHS too. That is: if $\omega$ is a root of $T(x^2)$, then $\omega^2$ is too. Also, if $\omega$ is a root of $T(x^4)$, then it's a root of $T(x^2)$, so $\sqrt{\omega}$ is also a root of $T(x^4)$. It follows (by looking at the argument of $\omega$) that $T(x^4)$ must have infinitely many roots, if it has one. Thus $T$ is a constant $d$, and $S(x)=dx-b$.

If $S(x)=dx-b$, then it's easy to check that $Q(x)=\frac{(x+b)^2}{d}-b$ works.

At last, $P(x)=Ax^{2^k}+B$, for reals $A, B$.

Sorry, I forgot the case $\omega=0$. The solution changes as follows: $T(x)$ can only have zero as a root, so $T(x)=d \times x^l$ for some $l$ and $S(x)=d \times x^{2l+1}-b$. Then $P(x)=d \times x^{(2l+1) \times 2^k}-b$, and $(2l+1) \times 2^k$ can take any positive integer value, so the working polynomials are just $P(x)=Ax^n+B$.

Just consider the second largest degree in both side.


For how many integers $ x$ is the following true:
$ |2\minus{}4x| < 14$.

hello, our inequality is equivalent with
$ \minus{}14<2\minus{}4x<14$
$ \minus{}16<\minus{}4x<12$
$ \minus{}3<x<4$
so all integer solutions are $ \minus{}2,\minus{}1,0,1,2,3$
Sonnhard.

Which means there are $ \boxed{6}$ integer solutions  :P

all you do is find the largest # and then the smallest # the smallest=-2 the largest=3 count how many in between the answer is [hide="small and big # answer"]6 is the # you we seeking [/hide][/quote]

[size=200][color=#FF4080]PLEASE RATE MY POST THANK YOU SO MUCH!!! IT IS GREATLY APRECIATED : D[/color][/size]

[size=50][color=#8080BF][b][i][u]I HAVE 99.58456656516165%CONFIDENCE THAT IM CORRECT [/u][/i][/b][/color][/size]

[color=#FF4040][b][i][u]PLEASE POST ME IF IM WRONG THANK YOU SO MUCH !!!![/u][/i][/b][/color]

 :gleam:  :gleam:  :gleam:  :gleam:


calculus
integration
trigonometry
geometry
Euler
function
special factorizations

Here is a question I have:
$\int x^{\frac{1}{2}}(1-x)^{\frac{1}{2}}dx$

I was thinking that substitution $x=sin^2\theta$ works but is there a simpler way?

It's not necessarily simpler, but here's something different:
$x-x^2=\frac14-\left(x-\frac12\right)^2$, so we substitute $x-\frac12=\frac12\sin\theta$

The difference between this and your idea is essentially a double-angle identity.

And if the question were about the definite integral

$\int_0^1\sqrt{x}\sqrt{1-x}\,dx,$

then I'd invoke geometry: this is the area of a semicircle, with center at $\left(\frac12,0\right)$ and radius $\frac12.$ (That's what you get from completing the square as jmerry did above.)

That would make this integral $\frac{\pi}8.$

Interesting historical tidbit:

John Wallis noticed the pattern, for nonnegative integers, that

$\int_0^1x^n(1-x)^n\,dx=\frac{(n!)^2}{(2n+1)!}$.

(That's not the notation he would have used, of course: this was [i]before[/i] Newton and Leibniz.)

He rather recklessly proclaimed, on the basis of the calculation I just did, that $\left(\frac12\right)!=\frac{\sqrt{\pi}}2.$ (Whatever $\left(\frac12\right)!$ is supposed to mean.)

Reckless, but after Euler developed the Gamma function, correct.

This is then a general case of the Beta integral:

$\int_0^1x^{\alpha-1}(1-x)^{\beta-1}\,dx=\frac{\Gamma(\alpha)\Gamma)(\beta)} {\Gamma(\alpha+\beta)}.$


parameterization
function
limit
real analysis
real analysis unsolved

Find all real parameters $ a$ for which function $ f: \mathbb{R}\rightarrow\mathbb{R}$ increases
$ f(x)\equal{}ax\minus{}ln(1\plus{}x^2)$

We must find all $ a$ such that $ \forall x \in \mathbb{R} \Rightarrow f'(x) > 0$. So we have:

$ a - \frac {2x}{1 + x^2} > 0$

$ a > \frac {2x}{1 + x^2} = \phi(x)\qquad \forall x \in \mathbb{R}$

Now we could try to find the maximum of $ \phi$ and set $ a$ to be grater that it:

$ \frac {d}{dx}\phi(x) = \frac {d}{dx}\left(\frac {2x}{1 + x^2}\right) = \frac {2}{1 + x^2} - \frac {4x^2}{(1 + x^2)^2} = 0$

$ 2(1 + x^2) - 4x^2 = 0$

$ x^2 = 1$

$ x_{1,2} = \pm 1$

Now:

$ \phi(1) = 1$

$ \phi( - 1) = - 1$

Now, since $ \phi(0) = 0$, $ \lim_{x \rightarrow \infty}\phi(x) = 0$ and $ \phi$ is continuous,  $ \phi(1) = 1$ is a maximum, so we can say that:

$ \forall a \in (1, + \infty) \Rightarrow f'(x) > 0$


Vieta
Gauss
number theory
relatively prime
complex numbers

Complex number $ z$ satisfies $ z^{2009}\equal{}1$.

Also, $ z^n$ never equals 1 for all $ n$ smaller than 2009.

Find the value of $ \frac{(1\plus{}z\plus{}z^2)(1\plus{}z^2\plus{}z^4)\cdots(1\plus{}z^{2008}\plus{}z^{4016})(1\plus{}z^{2009}\plus{}z^{4018})}{3}$.

Where are these problems you're posting lately coming from? They're nice problems.

They are from Korean Mathematics Olympiad for High Schoolers. (of course, not on-going)

This guy looks fun.

[hide]Note that $ 1 \plus{} z \plus{} z^2 \equal{} (z \minus{} \omega)(z \minus{} \omega^2)$, where $ \omega^3 \equal{} 1$ and $ \omega \neq 1$. The desired product is now just $ (1/3)\prod_{k \equal{} 1}^{2009 }(z^k \minus{} \omega) \prod_{k \equal{} 1}^{2009} (z^k \minus{} \omega^2)$. For the first half of the product, we note that $ z^k$ with $ 1 \leq k \leq 2009$ are the roots of $ w^{2009} \minus{} 1$, and the desired number is the product of all $ z^k \minus{} \omega$ with $ 1 \leq k \leq 2009$. By Vieta's formulas, we have that the desired product is the opposite of the constant coefficient of $ (w \plus{} \omega)^{2009} \minus{} 1$, so the product is $ \minus{} (\omega^{2009} \minus{} 1)$. Likewise, the other product is simply $ \minus{} (\omega^{4018} \minus{} 1)$. Multiplying this together, we get that our answer is $ (1/3)(\omega^{2009} \minus{} 1)(\omega^{4018} \minus{} 1) \equal{} (1/3)(\omega^2 \minus{} 1)(\omega \minus{} 1) \equal{} 1/3$. [/hide]

I think I did something wrong. <_<

Would you mind pointing out what my mistake was?

EDIT: Or nope, I didn't do anything wrong?

Your answer is wrong  :o

And I do not have the solution, just the answer...

By the way, the answer is 0.

[hide="Mieh..."]
Consider this product: $ (z \minus{} 1)(z^2 \minus{} 1)(z^3 \minus{} 1)\cdots (z^{2008} \minus{} 1)$. We know that it is not zero because it is stated that $ z$ is a primitive 2009th root of unity. Multiplying the expression in the problem by this product gives:
\[ \frac 13(z^3 \minus{} 1)(z^6 \minus{} 1)(z^9 \minus{} 1)\dots(z^{6021} \minus{} 1)(z^{6024} \minus{} 1)(1 \plus{} z^{2009} \plus{} z^{4018})
\]
We know that $ 1 \plus{} z^{2009} \plus{} z^{4018} \equal{} 1 \plus{} 1 \plus{} 1 \equal{} 3$, so the expression is simply:
\[ (z^3 \minus{} 1)(z^6 \minus{} 1)(z^9 \minus{} 1)\dots(z^{6021} \minus{} 1)(z^{6024} \minus{} 1) \\
\equal{} (z \minus{} 1)(z^2 \minus{} 1)(z^3 \minus{} 1)\dots(z^{2008} \minus{} 1)
\]
Since this is equivalent to the product we made above, we can only conclude that the original expression is equal to $ 1$.
[/hide]

Edit: Is the answer in fact 1?

Whoops, my apologies.

The original problem was when $ z^{2007}\equal{}1$, but I changed it to 2009, so the answer changed... :blush:

Wait if it is not zero then it must be $ 1$ (see my solution)?

I'm not sure. I'm seeing nothing wrong either my nor your solution, though I'm going to guess (not out of politeness, but genuinely) that my solution is incorrect.

EDIT: What do you know.
[quote="Zhero"]$ (1/3)(\omega^2 \minus{} 1)(\omega \minus{} 1) \equal{} 1/3$. [/quote]
Final step wrong. How disappointing. I apparently can't even plug 1 into the first formula I wrote down.

EDIT2: And Yongyi, your solution is prettier than mine. I need to stop bashing.

Sorry Zhero. I didn't catch that either. Your method was correct up until the last step. I didn't even pay attention to the value.

[hide="My solution was similar to yours Zhero"]
I noted that $ \prod_{k \equal{} 1}^{2008} (1 \plus{} z^k \plus{} z^{2k}) \equal{} \prod_{k \equal{} 1}^{2008} \left(\frac {z^{3k} \minus{} 1}{z \minus{} 1}\right)$
Since 3 is relatively prime to 2009, the k complex numbers $ z^{3k} \equal{} (z^k)^3$ are just a permutation (i.e. one-to-one rearrangement) of the k complex roots of unity, $ z^k$ (because the map $ a \to 3a$ is one-to-one in mod 2009).

So the product of the numerator is equal to the product of the denominator in the last expression, and the value of the expression is 1.[/hide]

[quote]My solution was similar to yours Zhero[/quote]
I think you mean Yongyi. :P

And it's OK, I think that was the solution that the problem writers were striving for. Mine was just an oh, look, this works! BOOM!

Thanks everyone!

One last question:

Yongyi, how did you conclude that $ (z\minus{}1)(z^2\minus{}1)\cdots(z^{2008}\minus{}1)\equal{}1$?

He didn't. See Gauss's post. Gauss is basically saying that the numerator and denominator are the same, so the product ends up being 1.

[quote="FantasyLover"]
Yongyi, how did you conclude that $ (z \minus{} 1)(z^2 \minus{} 1)\cdots(z^{2008} \minus{} 1) \equal{} 1$?[/quote]

I originally multiplied the expression by $ (z \minus{} 1)(z^2 \minus{} 1)\cdots(z^{2008} \minus{} 1) \equal{} 1$. Once I manipulated the resulting expression to match this one, it logically follows that the original one can only be equal to 1.


inequalities
LaTeX
inequalities proposed

[quote="yptsoi"]$a_{1},...,a_{n}$ are positive real numbers. Prove that
\[ a_{1} + \frac{2}{\frac{1}{a_{1}}+\frac{1}{a_{2}}} + \frac{3}{\frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}} + ... + \frac{n}{\frac{1}{a_{1}}+\frac{1}{a_{2}}+...+\frac{1}{a_{n}}} < 2(a_{1}+a_{2}+...+a_{n}). \][/quote]

This inequality was mentioned, with $\frac{1}{a_i}$ instead of $a_i$, in post #2 of the thread http://www.mathlinks.ro/Forum/viewtopic.php?t=18542 and http://www.mathlinks.ro/Forum/viewtopic.php?t=18543 . It is due to Hardy.

  Darij


On the Saturday of a weekend softball tournament, Team A plays Team B and Team C plays Team D. Then on Sunday, the two Saturday winners play for first and second places while the two Saturday losers play for third and fourth places. There are no ties. One possible ranking of the team from first place to fourth place at the end of the tournament is the sequence ACBD. What is the total number of possible four-team ranking sequences at the end of the tournament?

The given solution is correct, but the reasoning is somewhat difficult.  A simpler solution results from the observation that the order of finish is uniquely determined by the outcomes of the four games (A vs. B, C vs. D, Winners game, Losers game).  Each game has two outcomes, therefore there are 2^4 possible unique outcomes to the four games.  Thus, there are 2^4 = 16 possible orders of the teams.

I agree with NicksDad. I came up with the answer quickly using the same reasoning he did. I was able to follow the reasoning in the given solution, but it took some effort.

I have done this question before. If you want to see the answer click [hide="here"]
2 choices for the winner each game, and 4 games. So the answer is
\[ 2 \times 2 \times 2 \times 2=\boxed{16} \]
[/hide]

Just wondering...I solved the problem a slightly different way. Please let me know if its a longer way...
[hide]We could just solve possible ways for A being first place and multiply that number 4 (for each letter being first)
WE could have
ADBC
ADCB
ACDB
ACBD
We can't have B second place because in this scenario, it lost against A. Making it play with the other loser, B can either be 3rd or 4th
[/hide]
Thanks in advance :)

I used a different way to solve the problem.
For first place, there are 4 possibilities (A,B,C,D).
For each first place, such as A, there are 2 possibilities for second place.  (If A is first place, B cannot be second place).
For each set of first place and second place, there are two teams left that can be third place, and then one team left that can be fourth place.
So the answer is $4\cdot2\cdot2\cdot1=16$

@pearlm thats what i did too, i was surprised i got the right answer!

[hide=Solution]Looking at the ranking sequence, we see that A and B cannot both win on Saturday, and so neither AB nor BA can be the first and second places. Similarly, CD and DC cannot be the third and fourth places. Thus, the first and second place may be (A or B) and (C or D) or vice versa. This makes 2 (for the order) $\times 2\times 2 = 8$ possibilities for the first and second places. Then the third and fourth places can be two possible arrangements of the losers. In total, this is $8\times 2 = \boxed{16}$ arrangements.[/hide]


I don’t know how many of you have real experience of IOI but well I think at least weeblie has.

So I need response on this question:
Say memory limit is 64MB. So is it possible to declare: long a[1000000000]???

I mean, if the array size is within Memory Limit can I declare the index of array as large as possible?

Mahbub

The array size here is clearly more that 64 MB. It is almost 4 GB!

OOPS!!!  :blush: I did not saw it! Sorry! But main intesion was to ask whether it is possible to declare the array as large as possible within ML. And well u gave me the answer indirectly!!!

Thanks! :D

But, remember to not "touch" the boundaries.

Even an "empty" program uses some memory. ;)

Yes, it's better to stay safely in limit and fail the largest test case than try to make it tight and throw a SegFault on every one of them.

Anyway, lately the IOIs have a habit of having a lot of the test cases much smaller than the maximum, which sometimes really discourages from coding a good and efficient solution. I still remember my friend complaining about last year, i think it was in the Hermes problem, where he did some interval trees hung on interval trees and he got 100, but a very simple and naive-approach solution (much slower) earned 90 ...

I think you are mistaking the task. It was not Hermes but Artemis that used Interval Trees.

Much people coded N^2logN using interval trees and got 90-100, I even know my teammate wrote N^3 and got 80. I wrote pure (absolutely) N^2 and had just 55. That was awful task.


superior algebra
superior algebra solved

$R$ is a ring with multiplicative identity.
$u\in R$ has at least two right inverse.
Show that $u$ has infinitely many right inverses.

=========================================

I can prove that the cardinality of the set of all right inverses of $u$ is same to the cardinality of $Ru-1+s$, where $s$ is a right inverse of $u$.
How to prove that $R$ is infinite?

I realized it soon after making this post.


Find all positive integers N such that the product 2005 x N has exactly six divisors

[hide]2005 has 4 divisors: 1, 5, 401, and 2005. To get 6, we need to multiply by 5 or 401, yielding:
1, 5, 401, 25, 2005, 10025 or the alternate with 401 (instead of 5).[/hide]


function
integration
calculus
topology
complex analysis
complex analysis unsolved

Show that we can define an analytic branch of the  function $ f(z) \equal{}$ $ \sqrt {z^2 \minus{} 1}$ in any region(connected and open set, not necessarily [b]simply[/b] connected!) with the property that $ 1, \minus{} 1$ lie in the same component of the complement of the region.
What are the possible values of $ \int_{c}\frac {1}{\sqrt {z^2 \minus{} 1}}dz$ , where $ c$ is a closed curve(or a cycle, ie a closed chain of arcs) in the region?

You can't define $ \log(z)$ in the region, but you can define $ \log(z^2\minus{}1)$. Once you have that there's a clear choice for a definition of $ f(z)$.

The integration part is independent of how you defined $ f$, conveniently. Doesn't $ \frac1{\sqrt{z^2\minus{}1}}$ look a lot like $ \frac{1}{z}$ when $ z$ is large? :)

Well, up to a $ \pm$ sign. Negating the function does negate the integral.

True, true. What I meant was that you don't need your construction in order to do the second part; you only need that a coherent analytic $ \sqrt{z^2\minus{}1}$ exists.

Also, it's $ \sqrt{1\minus{}z^2}$ in the book.


18. You are given a vector in the xy plane that has a magnitude of 90.0 units and a y component of -55.0 units.

(a) What are the two possibilities for its x component?

(b) Assuming the x component is known to be positive, specify the vector which, if you add it to the original one, would give a resultant vector that is 80.0 units long and points entirely in the -x direction.

[Please help me with these. Explain clarely if possible.]

I thought you'd remember that Pyhagorean Theorem by now, airelemental :) . No, sorry, no offence. Anyway:

$($magnitude of vector$)^2$ = $($x component$)^2$ + $($y component$)^2$

So if you know that the magnitude is 90 and y component is -55, you get:

$90^2=x^2+(-55)^2$

$x^2=5075$

$x=\pm71.24$

Now for (b), we take x component = 71.24, right? The vector we're looking for has some componenets $a$, $b$. If you add this vector to the original one, you get vector with components

$71.24+a$

and

$-55+b$.

You want the result to be entirely in -x, so the result has zero y component. If it has zero y component, it's magnitude is

$80^2 = ($x component$)^2 + 0^2$
$\pm80=$ x component

And it has to be -80, as we know. So the vector we're supposed to get is (-80, 0). Therefore:

$71.24+a=-80$
$a=-151.24$

$-55+b=0$
$b=55$


calculus
integration
function
calculus computations

For a function $ f(x) \equal{} 6x(1 \minus{} x)$, suppose that positive constant $ c$ and a linear function $ g(x) \equal{} ax \plus{} b\ (a,\ b: \text{constants}\,\ a > 0)$ satisfy the following 3 conditions: $ c^2\int_0^1 f(x)\ dx \equal{} 1,\ \int_0^1 f(x)\{g(x)\}^2\ dx \equal{} 1,\ \int_0^1 f(x)g(x)\ dx \equal{} 0$. Answer the following questions.

(1) Find the constants $ a,\ b,\ c$.

(2) For natural number $ n$, let $ I_n \equal{} \int_0^1 x^ne^x\ dx$. Express $ I_{n \plus{} 1}$ in terms of $ I_n$. Then evaluate $ I_1,\ I_2,\ I_3$.

(3) Evaluate the definite integrals $ \int_0^1 e^xf(x)\ dx$ and $ \int_0^1 e^xf(x)g(x)\ dx$.

(4) For real numbers $ s,\ t$, define $ J \equal{} \int_0^1 \{e^x \minus{} cs \minus{} tg(x)\}^2\ dx$. Find the constants $ A,\ B,\ C,\ D,\ E$ by setting $ J \equal{} As^2 \plus{} Bst \plus{} Ct^2 \plus{} Ds\plus{}Et \plus{} F$.
(You don't need to find the constant $ F$).

(5) Find the values of $ s,\ t$ for which $ J$ is minimal.

Hello, Kabi. Your answers are correct so far.

Kunny, u have forgotten to write D in question $ 4$.I hope it is coefficient of $ s$.
[hide="4"]
$ J \equal{} \int_0^1 (e^x \minus{} s \minus{} t(2\sqrt {5}x \minus{} \sqrt {5}))^2\ dx$
$ A \equal{} 1,B \equal{} 0,C \equal{} \frac {5}{3},D \equal{} 2(1 \minus{} e),E \equal{} 2\sqrt {5}(e \minus{} 3)$
[/hide]
First tell me whether i have calculated anything wrong or not.
Thank u

Sorry, I made a typo. :oops:

I have just edited.

regards,

kunny


$\lim_{n\to\infty}s_n=\infty.$

No, it doesn't converge.

Clearly, each of $(a_n),(b_n)$ and $(c_n)$ is an increasing sequence. Thus, they are subject to the "monotone sequence alternative": either they converge or they tend to infinity.

Suppose each of them converges: $a_n\to A,b_n\to B, c_n\to C.$ Then, taking the limit in the recursive formula,

$A=A+\frac1{2^{2548}(B+C)}.$

This is impossible. Hence at least one of the sequences tends to infinity and $s_n\to\infty.$

(Slowly!)


geometry
inradius
trigonometry
circumcircle
incenter
ratio
geometric transformation

Consider the feet of the orthogonal projections of $ A,B,C$ of triangle $ ABC$ onto the external angle bisectors of angles $ BCA, CAB$, and $ ABC$, respectively. Let $ d$ be the length of the diameter of the circle passing through these three points. Also, let $ r$ and $ s$ be the inradius and semiperimeter, repectively, of triangle $ ABC$. Prove that $ r^{2}\plus{}s^{2}\equal{}d^{2}$.

.........

let's put $ A_1$ be a perpendicular foot from $ A$ to the external bisector of $ \angle BCA$
            $ A_2$ be a perpendicular foot from $ A$ to the external bisector of $ \angle ABC$

put $ B_1, B_2, C_1,C_2$ simialrly....

and more, put $ A'$ be a excenter of $ A$. $ B', C'$ are similarly defined.

Claim) $ A_1,A_2,B_1,B_2,C_1,C_2$ are concyclic.

it is pretty obvious that $ A_2, B_1, A, B$ are concyclic.
so $ C'B_1*C'A \equal{} C'A_2*C'B$.
meanwhile, $ BC_1 \equal{} a\sin\dfrac{B}{2}$ and $ AC_2 \equal{} b\sin\dfrac{A}{2}$
so, $ \dfrac{BC_1}{AC_2} \equal{} \dfrac{a\sin\dfrac{B}{2}}{b\sin\dfrac{A}{2}} \equal{} \dfrac{\cos\dfrac{A}{2}}{\cos\dfrac{B}{2}} \equal{} \dfrac{C'B}{C'A}$
so, from  $ C'B_1*C'A \equal{} C'A_2*C'B$, we can obtain $ C'A_2*C'C_1 \equal{} C'B_1*C'C_2$.
therefore, $ A_2, C_1, C_2, B_1$ are concyclic.
similarly, $ B_1, C_2, A_1, B_2$ and $ A_1, B_2,C_1,A_2$ are concyclic too.

if $ A_1,A_2,B_1,B_2,C_1,C_2$ are not concyclic,
radical axis of the circumcircle of quadrangle$ A_2 C_1 C_2 B_1$ and the circumcircle of quadrangle$ B_1 C_2 A_1 B_2$ and
radical axis of the circumcircle of quadrangle$ A_2 C_1 C_2 B_1$ and the circumcircle of quadrangle$ A_1 B_2 C_1 A_2$ and
radical axis of the circumcircle of quadrangle$ B_1 C_2 A_1 B_2$ and the circumcircle of quadrangle$ A_1 B_2 C_1 A_2$ are
concurrent.
it means $ B_1C_2$, $ A_1B_2$,$ C_1A_2$ are concurret, which is contradiction.
so $ A_1,A_2,B_1,B_2,C_1,C_2$ are concyclic.
-------------------------------------------------------------------------------------------------------------------------------------------------

now then $ d$ becomes the diameter of the circumcircle of hexagon$ A_1 A_2 B_1 B_2 C_1 C_2$

meanwhile, $ A_1C_1^2 \equal{} CC_1^2 \plus{} CA_1^2 \minus{} 2CC_1*CA_1*\cos\angle A_1CC_1$
because $ \dfrac{A_1C_1}{\sin\angle A_1B_1C_1} \equal{} d$ and $ \angle A_1B_1C_1 \equal{} \angle A'B_2C_1 \equal{} \dfrac{\pi}{2} \minus{} \dfrac{B}{2}$,
$ d \equal{} \dfrac{A_1C_1}{\cos\dfrac{B}{2}}$
so $ d^2 \equal{} \dfrac{A_1C_1^2}{{\cos\dfrac{B}{2}}^2}$
with some calculating, we can obtain that $ d^2 \equal{} s^2 \plus{} \dfrac{(s \minus{} a)(s \minus{} b)(s \minus{} c)}{s}$
and from $ sr \equal{} \sqrt {s(s \minus{} a)(s \minus{} b)(s \minus{} c)}$, $ r \equal{} \sqrt {\dfrac{(s \minus{} a)(s \minus{} b)(s \minus{} c)}{s}}$

so $ d^2 \equal{} s^2 \plus{} r^2$.

QED

[quote="limsk1"]with some calculating, we can obtain that $ d^2 \equal{} s^2 \plus{} \dfrac{(s \minus{} a)(s \minus{} b)(s \minus{} c)}{s}$[/quote]
Since I was not good at trigonometric calculation, I could not calculate $ A_{1}C_{1}^{2}$.............
But I solved this problem using different method as yours..
Define point similary. Let $ I$ be a incenter of triangle $ ABC$, The incircle of triangle $ ABC$ touches $ BC,CA,AB$ at $ D,E,F$ repectively, and $ A$-excircle touches $ AC$ at $ G$.
$ X \equal{} BA'\cap C_{2}C$. then it is easy to show that $ CA_{1}A \sim CC_{1}X$, Therefore $ CA_{1}C_{1} \sim CAX$.
$ \angle CC_{1}A_{1} \equal{} \angle CXA \equal{} \theta$.
( we can easily show that $ B_{1}C_{1} \parallel AX$ and $ \angle CC_{1}A_{1} \equal{} \angle BB_{1}C_{1}$, but we don't needs this property.)
I conjectured that $ A_{1}B_{1}C_{1}\sim DEF$. to show this and calculate its ratio, It is enough to show that
$ \left(\frac {2A_{1}C_{1}}{DF}\right) ^{2} \equal{} 1 \plus{} \frac {s^{2}}{r^{2}}$.

but $ \frac {A_{1}C_{1}}{\sin \frac {A}{2}} \equal{} \frac {CA_{1}}{\sin \theta}$ and $ DF \equal{} 2r\cos \frac {B}{2}$,
$ \frac {2A_{1}C_{1}}{DF} \equal{} \frac {CA_{1}\sin\frac {A}{2}}{r\cos\frac {A}{2}\sin\theta} \equal{} \frac {b \sin\frac {C}{2}\sin\frac {A}{2}}{r\cos\frac {B}{2}\sin\theta} \equal{} \frac {4R\sin\frac {A}{2}\sin\frac {B}{2}\sin\frac {C}{2}}{r\sin\theta} \equal{} \frac {1}{\sin\theta}$.
Hence $ \left(\frac {2A_{1}C_{1}}{DF}\right) ^{2} \equal{} 1 \plus{} \cot^{2} \theta$.
so remains to prove is $ \cot \theta \equal{} \frac {s}{r}$.

It can be proved as follows...
$ \cot \theta \equal{} \frac {XC_{2}}{AC_{2}} \equal{} \frac {XC_{2}}{CC_{2}}\frac {CC_{2}}{AC_{2}} \equal{} \frac {AA'}{AI}\frac {AE}{IE} \equal{} \frac {AG}{AE}\frac {AE}{IE} \equal{} \frac {AG}{IE} \equal{} \frac {s}{r}$.
($ ACC_{2}\sim IAE$, $ AI\parallel C_{2}X$, $ IE\parallel A'G$)

I wonder that calculation can simpler as this....
Anyone has a solution does not contains calculation like this?

$ I_{A},I_{B},I_{C}$ =excenter.

$ D,E,F$ = midpoint of $ BC,CA,AB$

$ X,Y,Z$ = orthogonal projections of $ A,B,C$ of triangle $ ABC$ onto the external angle bisectors of angles $ BCA, CAB$, and $ ABC$

$ K$ = incenter of $ DEF$.

$ P$ = orthogonal projections of $ A$ onto external angle bisectors of angles $ ABC$

I will prove $ K$ is circumcenter of trianlge $ XYZ$.
Since triangle $ ABC$ and triangle $ DEF$ are homothety, $ EK\parallel BI$ and $ BI\perp I_CI_A$. So $ EK\perp I_CI_A$. Also, in quadrilateral $ ZCAP$, it is easy to see $ EZ \equal{} EP$. Together with fact that $ EK\perp I_CI_A$, we obtain $ KZ \equal{} KP$. In triangle $ I_CAB$, $ F$ is midpoint of $ AB$ and $ P,Y$ are projection of $ A,B$. Hence $ FP \equal{} FY$. To prove $ KP \equal{} KY$, we have to show $ KF\perp PY$. By homothety, $ KF\parallel CI$ and $ CI\perp PY$ hold because  $ I_{C}I$ is diameter passing $ C$ of triangle $ I_CAB$ and $ PY$ is antiparallel to $ AB$. Hence $ KP \equal{} KY$ and from $ KZ \equal{} KP$, we get $ KZ \equal{} KY \equal{} KP$. Similarly, $ KZ \equal{} KX$. So $ K$ is circumcenter of circle that pass $ X,Y,Z,P$.
$ C$ is orthocenter of triangle $ II_BC$. and $ E$ is midpoint of segment that joining $ A$ and orthocenter of triangle $ II_BC$. Hence Simson line of $ A$  wrt $ \triangle II_BC$ pass $ E$ and also pass $ X$. ie, $ XE$ is Simson line of $ A$ wrt $ \triangle II_BC$. Similarly, $ PF$ is Simson line of $ A$ wrt $ \triangle II_CB$. And It is obvious that Simson line of $ P$ wrt $ \triangle II_BC$ and wrt $ \triangle II_CB$ are coincide. Hence $ X,E,F,P$ are collinear. Let incircle of $ DEF$ meet $ EF$ at $ T$.  Then $ T$ is midpoint of $ XP$(since $ KX \equal{} KP$) and $ KP \equal{} \frac {d}{2}, KT \equal{} \frac {r}{2}$(by homothety). $ KP^2 \equal{} KT^2 \plus{} PT^2$ become $ d^2 \equal{} r^2 \plus{} XP^2$. So it is suffcie to show $ XP \equal{} s$ which is hold since $ XP \equal{} XE \plus{} EF \plus{} FP \equal{} \frac {CA}{2} \plus{} \frac {BC}{2} \plus{} \frac {AB}{2} \equal{} s$.


Just for knowing what is the definition of a parabola because i got into an argument with my dad over it, he said that its only made by the $x^2$ equation.  I think that there are others.  Could someone help answer this for me

Here is the primitive definition of parabola:
[url=http://mathworld.wolfram.com/Parabola.html]mathworld.wolfram.com/Parabola.html[/url]

Standard forms: $y=ax^2+bx+c$, $y-h=(x-k)^2$. Any polynomial of degree 2.

ok so its only 2nd degree polynomials.   Thx guys


inequalities
number theory unsolved
number theory

Let $d ( n )$ be the number of divisors of $n$. Find all $n$ satysfying
\[ d ( n + d ( n ) ) = d(n)  \]
An exemple is $n = 12$, since $d ( 12 ) = 6$ and $d ( 12 + 6 ) = 6$.
I haven't give it much tought, and didn't solve it ;(. But I guess it is solvable (doesn't look too hard) so I put it in this section.

I don't think $n = 12$ works :)

Assuming that your equation is $d(n + d(n)) = n$: it is not hard to get a bound on $d(n)$ in terms of $n$. One of the possible estimates (quite a weak one) is $d(n) \leq 2\sqrt{n}$. Hence we get \[ n = d(n + d(n)) \leq 2\sqrt{n + 2\sqrt{n}}  \] and from here it is easy to get a bound on $n$. Then you just need to check cases.

Or did you mean $d(n + d(n)) = d(n)?$ :) In that case, $n = 12$ does work...

Yes, I meant $d(n+d(n))=d(n)$, I'm really sorry ...
Hope I didn't waste too much brainpower of mathlinkers ;)

Well, if $n$ and $n+2$ are twin primes, then $d(n+d(n)) = d(n+2) = 2 = d(n)$, so describing the complete set of solutions for your equation could be quite complicated...

Yes, I know. Homever I'm still not discouraged. (maybe because I didn't give it enough tought still ;))[/hide]

No solution, just some more results... This problem is quite challenging! Well, I'm going to prove $d(n) \equiv 0 \bmod 2$. Suppose the contrary. Then a fortiori $n = a^2$, for some $a \in \mathbb{Z}^+$, and $d(n+d(n)) = d(n)$ only if $d(a^2 + d(a^2)) \equiv 1 \bmod 2$. Once more this means $a^2 + d(a^2) = b^2$, for any integer $b > a$. Anyway $a^2 < a^2 + d(a^2) \leq a^2 + 2a < (a+1)^2 \leq b^2$, where we made use of the inequality that Arne already mentioned above. This leads to an absurd, forcing $d(n)$ to be even.

Maybe I'll renounce: I just found $d(5^3) = 4 = d(3 \cdot 43) = d(5^3 + d(5^3))$.


Let's see...

Kevin Chen

11 other people who I don't know...

Kevin Chen
Bobby Shen
Allen Yuan (why did he get second at state?)
Carl Lian
Richard Ni
Wenyu Cao (why did he get second at state?)
Mario Choi
Abubakar Abid
Preet Patel (why did he get third at state?)
Lucy Chen
Symon Gray
Shijie Joy Zheng

you guys are saying "why did he get 3rd at state?"

remember Albert NI? he got 2nd at school and 3rd at state. went on to win nationals.

Go Preet!
I can't predict as I don't know what these people plan to do before nationals.

texan pride XD
kevin + bobby = sheer pwnage! =D

Richard Ni's gonna be up there for sure.

Unfortunately, nothing is "for sure" in mathcounts.

He's VERY LIKELY to be up there. Good call.
(I hate probability... and it hates me, too)

Yeah right. My parents say if I make it to countdown they'll buy me a computer of my choice and a projector screen...  :rotfl:

I have a decent chance of making it in the top 100. Don't count on me being in the top 12 (haha... "count"... I'm so funny. Ignore me.)

No particular order:

Kevin Chen
Jonathon Horton
Allen Yuan
Carl Lian
Ben Alan White
Wenyu Cao
Mario Choi
InYoung Cho
Preet Patel
Lucy Chen
Shijie Joy Zheng
oh who am i kidding i vote meself in ( I voted all the other 3-timers in :maybe: )

Kevin Chen
Wenyu Cao
Allen Yuan
Bobby Shen
Carl Lian
Preet Patel
Shijie Zheng
Lucy Chen
(no particular order)
idk after that...

Kevin Chen
Allen Yuan
Bobby Shen
And some other people

[quote="Fanatic"]LOL @ it hates me too

I hate geometry and vice versa, if it makes you feel better xP.

Anyhow, in no order [I have numbers to keep track for my own good]

1. Kevin Chen
2. Carl Lian
3. Mario Choi
4. Allen Yuan
5. Someone I don't know
6. Someone I don't know
7. Someone I don't know
8. Someone really good that I'm forgetting at the moment
9. Someone from California
10. Elizabeth Synge
11. Joy Zheng
12. Kevin Hu (I have to put my last year's predicted winner pick in the top 12!)

Btw, i'm not predicting myself to be either in or out of the T12, just leaving myself out completely.[/quote]

oh man this kid knows what's up!
go Carl Lian!

but here's mine, revised from before.

1. Kevin Chen
2. Wenyu Cao
3. Allen Yuan
4. Carl Lian
5. Mario Choi
6. Lucy Chen
7. Elizabeth Synge
8. Shijie Zheng
9. Preet Patel
10. Bobby Shen
11. Richard Ni
12. Ching Chong

I believe that Jonathan Horton shall be. I don't know anyone else well enough.

Also, let's stick to the topic. It is good for your post count to stick to the topic.

ya billy! i picked jonathon horton too. he's a nice kid.

I predict that some of them were in Illinois at least once in their life. :)

Why is everyone betting on these things? It's unpredictable,  like in 2005, when Neal Wu somehow won.

Fanatic's brother, he won in 2002? as a 7th grader.

No he was an 8th grader when he won.

You're confusing him with Ryan Ko, who won in 7th grade

i think shijie zheng should be higher, maybe 5th or so.
and what about daniel li? unless i just have TERRIBLE eyesight like i think i do. oh well.
bobby shen
a lot of kevins (lol)
what grade is silvis in?
lucy chen
mario choi
elizabeth synge
what grade is iris xu in?

and others that im 2 lazy 2 list...
i just know it wont be me.

Oh, and i think a georgian will NOT make it. we're pretty horrible.

Li and Silvis are both in 9th now, unless I'm being very stupid, espicially since I was on the same team as silvis last year...


The radius of a circle is decreased by $ 20\%$. By what percent is its area decreased?

$ r \equal{}$ original radius, then $ r\minus{}.2r \equal{} .8r$ is decreased radius

$ r^2 \pi \equal{}$ original area, then $ .64r^2\pi$ is decreased area.

(original area - decreased area)/original area = .36 = 36%

So the answer is [b]36%[/b].

EDIT: lol stupid me..

Hmm, I'm not sure what you did, but I this is what I got:

Let $ r$ be the original radius.  The area of the circle is $ r^2\pi$.  The decreased radius is $ r\minus{}.2r\equal{}.8r$.  So the area of the smaller circle is $ \pi (.8r)^2\equal{}.64r^2\pi$  Therefore, the amount of decrease is \[ \frac{r^2\pi\minus{}.64r^2\pi }{r^2\pi }\equal{}\frac{.36r^2\pi}{r^2\pi}\equal{}.36\equal{}\boxed{36\%}\]

I did it like this:

We know that the area of a circle is $ \pi r^2$. So, we know that r is $ 80 \%
$. So, we can say  $\frac{6400}{10000}$. Since that is $r^2$. We then find it is $64%
$. So, we can find the number of subtraction to be $ \boxed{36\%}
$.


calculus
integration
function
limit
real analysis
real analysis unsolved

If $ \int_{0}^{\infty}f(x) dx$ converges, and $ f$ is uniformly continuous on $ [0,\infty)$, then $ \lim_{x\to\infty}f(x) \equal{} 0$.

Can you give me any hints on this?

Thanks.

If $ |f(x)\minus{}f(y)|\le\epsilon$ whenever $ |x\minus{}y|\le\delta$, then $ \left|\int_{x\minus{}\delta}^{x\plus{}\delta}f(t)\,dt\right|\ge 2\epsilon\cdot\delta$ whenever $ |f(x)|\ge 2\epsilon$.

Infinitely many of these blocks breaks convergence.


modular arithmetic
symmetry
rotation
counting
combinatorics
graph theory
IMO Shortlist

A game is played by $n$ girls ($n \geq 2$), everybody having a ball. Each of the $\binom{n}{2}$ pairs of players, is an arbitrary order, exchange the balls they have at the moment. The game is called nice [b]nice[/b] if at the end nobody has her own ball and it is called [b]tiresome[/b] if at the end everybody has her initial ball. Determine the values of $n$ for which there exists a nice game and those for which there exists a tiresome game.

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

$n$ is nice iff $n=2$ or $n \ge 4$:

It is easy to see that $n=1,3$ are not nice and $n=2$ is.

Let $t(a,b)$ denote that person $a$ and $b$ traded balls.  Thus, if we do $t(1,2)\cdots t(1,n)$ (from left to right) then we have that person $k$ has ball $k-1 \pmod n$.  If we do $t(n,n-1)\cdots t(n,2)$ then we have that $1$ and $n$ exchanged ball and everyone else has their own ball.  Thus, since we have done all exchanges using $1$ or $n$ and have done no exchange that do not have $1$ or $n$, we can conclude that if $n-2$ is nice then $n$ is.  We can easily find a series of exchangements that show that $n=5$ is nice $(t(1,2)t(1,3)t(1,4)t(1,5)t(2,3)t(2,4)t(2,5)t(3,5)t(3,4)t(4,5))$; therefore, all evens $\ge 2$ and all odds $\ge 5$ are nice.

$n$ is tiresome iff $n\equiv 0,1\pmod 4$.

To prove the 'only if' part, note that if $n\equiv 2,3\pmod 4$ then $\binom{n}{2}$ is odd.  Since the exchanges are isomorphic to the transpositions of $S_n$, we have the composition of all of the transpositions is a symmetry of odd sign, but the identity has even sign, contradiciton.  (See: [url]http://en.wikipedia.org/wiki/Sign_of_a_permutation[/url])

To prove the 'if' part, consider the case that $n\equiv 0\pmod 4$, do $t(k,k+n/2)$ for $n/2\ge k\ge 1$. Notice how this rotates each of balls $1$ through $n$ $180$ degrees. Now for $k=1$ to $k=n/2$, do $t(k,k+1)t(k+1,k+2)\cdots t(k-1+n/2,k)$ and $t(k+1+n/2,k+2+n/2)\cdots t(k-1,k+1+n/2)$, this will rotate each time $1$ through $n$  $360/n$ degrees except $n/2$ and $n$ which are each rotated $180+360/n$ degrees.  After $n/2$ times, each of $1$ through $n$ except $n/2$ and $n$ are rotated $180$.  $n/2$ and $n$ are rotated $180(n/2)+180=180$ degrees.  Thus, all of $1$ through $n$ is rotated $180+180=360=0$ degrees i.e. $n$ is nice.
To prove that this uses all $\frac{n(n-1)}{2}$ transpositions, notice that the first part uses $n/2$ transpositions and the second part uses $(n-2)n/2$ transpositions which sums to $(n-1)n/2$, as desired.

If $n\equiv 1\pmod 4$ do $t(1,2)t(2,3)t(1,3)\cdots t(1,n-1)t(n-1,n)t(1,n)$.  Thus if we remove $1$, we are in a position equivalent to the second part of the case where $n\equiv 0\pmod 4$, so we are done, as desired.

[b]Part 1: Niceness[/b]
Denote the balls by $a_1,a_2,\dots, a_n$ and we wish to find $n$ such that we can transpose each pair of positions, one at a time, and get no fixed points. We claim that $n$ is nice if and only if $n\neq 3$.

Indeed, if $n=3$ and if WLOG positions $(a,b)$ switch and then positions $(b,c)$ switch, then positions $(c,a)$ must switch, which leaves $a$ with the same ball as before.

Note that we can do the following transpositions: $(i,i+1), (i,i+2), \dots (i,n)$ and we will do a cyclic shift, turning $(i,i+1,\dots, n)$ into $n,i,i+1,\dots, n-1.$ Call this sequence of moves $C_1(i).$ If we do these same transpositions in the opposite direction, we will get a different cyclic shift: $i+1,i+2,\dots, n, i.$ Call this sequence of moves $C_2(i).$

If we do $C_1(1)$ and then $C_2(n)$ then we will get rid of the need to make any operations on position $1$ or $n$, as well as get the same configuration as before except for $1$ and $n$ transposed. Therefore, if $n-2$ is nice then $n$ is nice. Note that $n=2$ is nice because $C_1(1)$ and $n=5$ is nice because $C_1(1), C_1(3), C_1(2), C_1(4)$ so our claim is proved.

[b]Part 2: Tiresomeness[/b]
Let $S$ be the number of pairs of balls $(a_i,a_j)$ such that $i<j$ and $a_i$'s position is to the right of $j$'s position. Originally $S=0$. Note that each transposition changes $S$ by $1$, so if the ending $S$ is $0$ there must be an even number of moves made, so $n$ is tiresome only if $\tbinom{n}{2}$ is even, or when $n\equiv 0,1\pmod 4.$

If $n=4$, then note that the sequence of transpositions $(1,2),(3,4), (1,4) ,(2,3) ,(1,3), (2,4)$ is enough. If $n=5$ we can find a construction easily.

We proceed by induction: if $4k$ is tiresome, then we add on girls $4k+1,4k+2,4k+3,4k+4$. We can make the first $4k$ finish all their internal transpositions and have them each retain her own ball and the last $4$ do the same. Note that $(4k+1,i),(4k+2,i+1),(4k+1,i+1),(4k+2,i)$ is equivalent to $(4k+1,4k+2),(i,i+1)$ and call this transformation $D(4k+1,i).$ Now, $D(4k+1,1),D(4k+3,1),D(4k+1,3),\dots, D(4k+3,4k-1)$ does the trick.

If $4k+1$ is tiresome, then we do $D(4k+2,1),D(4k+4,1),\dots, D(4k+3,4k-1)$ and then the tiresome transposition sequence of $4k+1$, and then do the tiresome transpositions sequence if $5$ on $4k+1$ to $4k+5$. We are done.

This problem should be COMBINATORICS :mad:


The sum of the two $ 5$-digit numbers $ AMC10$ and $ AMC12$ is $ 123422$. What is $ A\plus{}M\plus{}C$?

$ \textbf{(A)}\ 10 \qquad
\textbf{(B)}\ 11 \qquad
\textbf{(C)}\ 12 \qquad
\textbf{(D)}\ 13 \qquad
\textbf{(E)}\ 14$

$ 123422 $ is $ 5 (mod 9)$ . Therefore $ 2 \cdot AMC $ is 1 (mod 9), and $ A+M+C=14$.

Setting this up:
AMC10
+AMC12
=
123422

This means $2C = 4$ or $14$, but we notice that $2M = 3, 13$, which is clearly impossible unless there is some composition. Meaning $C = 7$. We notice that $2A = 2, 12$, but if there is any composition, then it would be an odd number, so $M = 1$. Next we note that $A + A = 12$, so $A = 6$. Thus these sum to $7+1+6 \rightarrow \boxed{14}$.


inequalities
algebra
polynomial
algebra proposed

$ c$ is a positive integer. Consider the following recursive sequence: $ a_1\equal{}c, a_{n\plus{}1}\equal{}ca_{n}\plus{}\sqrt{(c^2\minus{}1)(a_n^2\minus{}1)}$, for all $ n \in N$.
Prove that all the terms of the sequence are positive integers.

[hide="How to be Lazy and Bring in the Artillery"]
Look [url=http://mathworld.wolfram.com/ChebyshevPolynomialoftheFirstKind.html]here[/url], particularly the recurrence relations and formula (17) for the Chebyshev Polynomials of the first kind, $ T_n$. The former shows you that these numbers are just $ a_n\equal{}T_n(c)$, and the latter tells you that these values are integers.

I'm sure this can be proved directly, but given the Chebyshev inspiration it's almost impossible to work on this problem without relying on that as a huge hint.
[/hide]


Would someone point me towards a website that gives a good explanation of how to factor in air resistance & friction when calculating projectile motion? Or, perhaps just give me some keywords to use, because google isn't giving me what I want with my current keywords.

Air resistance is hard to calculate precisely. Usually the air resistance (resistance force) is calculated as $-cv^2$ or with some other exponent, see Reynolds number. If you're interested in calculating the trajectory of a bullet or some other real-world object, the best way to do it is to actually measure all the parameters. Specifically for bullets, google for ballistic tables, also spheres are somwhat easier to calculate, but simulating a falling piano might be a bit of a challenge...


Two noncongruent circles $k_1$ and $k_2$ are exterior to each other. Their common tangents intersect the line through their centers at points $A$ and $B$. Let $P$ be any point of $k_1$. Prove that there is a diameter of $k_2$ with one endpoint on line $PA$ and the other on $PB$.

Let $CD$ be the diameter of $k_2$ which is parallel to $O_1P$ (where $O_1,O_2$ are the centers of $k_1,k_1$ respectively). $A,B$ are the centers of the two similarities which turn the circles into one another, one internal and one external. One of these similarities will take $C$ onto $P$ (because $C,P$ are endpoints of diameters which correspond to one another through the two similarities), while the other one will take $D$ onto $P$, and the conclusion follows.


limit
number theory
real analysis
real analysis unsolved

Does exists $x\in Q$, $x>1$ such that $\exists \ \lim_{n\to\infty}\{x^n\}?$ ($\{a\}=a-[a]$)

OK... but excluding this trivial case..

Denote $a_n=\{x^n\}$ and $b_n=[x^n]$, $x=\frac p q$ and $m_n=(p\cdot b_n) \text{mod q}$ and $l$ be the limit of $a_n$ as $n$ goes to infinity.

Let's take the case in which $\frac p q l$ is not an integer.

We have $a_{n+1}=\{\{\frac p q a_n\}+\{\frac{m_n} {q}\}\}$ tending n to infinity we have:
$l=\{\{\frac p q l \}+\{\frac{m_n} {q} \}\}$ (1) and so $m_n$ could take a single value. Denote it by m.
We have also
\[ b_{n+1}=[\frac p q b_n+\frac p q a_n]=\frac{pb_n-m}{q}+[\frac p q a_n] +[\{\frac p q a_n\}+\{\frac m q\} ]  \]
tending n sufficiently large we obtain:
\[ b_{n+1} =\frac{pb_n-m}{q}+[\frac p q l]+[\{\frac p q l\}+\{\frac m q \} ]=\frac{pb_n-m}{q}+[\frac p q l]+c  \]
where c  is equal to $[\{\frac p q l\}+\{\frac m q \} ]$  (2). Now denote $c+[\frac p q l]=s$(3).
We have $q(b_{n+1}-s)=(pb_n-m),\forall n>N$(4).Where N is a fixed integer sufficiently large.

Now we have $(b_n+a_n)\frac p q=b_{n+1}+a_{n+1}=\frac{pb_n-m}{q} +s+a_{n+1} \Rightarrow$
$a_n\frac p q= s- \frac m q +a_{n+1}\Rightarrow \frac{p-q} {q} l=s- \frac m q\Rightarrow l=\frac{sq-m}{p-q}$.

I think this case is simpler :D  From $a_{n+1}=\{\{a_n \frac p q\}+\{m_n\frac p q\}\}$ tending n to infinity we have $l=\{m_n \frac p q \}$ because $\{a_n \frac p q\}$ goes closer to 1 or to 0 as n grows. Now we conclude that $m_n$ can take a single value, say it  $m$. And $l=\frac{mp-kq}{q}$ but $l \frac p q= \frac{p(mp-kq)}{q^2}$ which cannot be integer if $q\neq 1$. Contradiction.
The only case when the sequence has limit is when $x\in \mathbb{Z}$. :P

xirti can i ask you some questions?

[quote]and $m_n=p\cdot b_n \text{mod q}$ [/quote]

what you mean? $m_n=(p\cdot b_n) \text{mod q}$ or $m_n=p\cdot (b_n \text{mod q})$. Or maybe you want only  $m_n= b_n \text{mod q}$ :?:

[quote] We have $a_{n+1}=\{\{\frac p q a_n\}+\{m_n\cdot \frac p q\}\}$  [/quote]
but $\{x^{n+1}\}=\{[x^n]\frac{p}{q}+\{x^n\}\frac{p}{q} \}=$
$=\{(wq+ [x^n] (mod\ q)) \frac{p}{q}+\{x^n\}\frac{p}{q} \}=\{ ([x^n] (mod\ q)) \frac{p}{q}+\{x^n\}\frac{p}{q} \}=$
$=(b_n (mod\ q)) \frac{p}{q}+\{x^n\}\frac{p}{q} \}$ so i think $\{(b_n (mod\ q)) \frac{p}{q}\}$ must be euqal to [quote]$\{m_n\cdot \frac p q\}$[/quote] and maybe you mean  $m_n= b_n \text{mod q}$.
If $m_n=p\cdot (b_n \text{mod q})$ i cant understand why \[ \{(b_n (mod\ q)) \frac{p}{q}\}=\{(b_n \text{mod q})\cdot \frac{p^2}{q}\}=\{p\cdot (b_n \text{mod q})\cdot \frac p q\} =\{m_n \frac p q\} \].
And if you mean $m_n=(p\cdot b_n) \text{mod q}$, then must be $a_{n+1}=\{\{\frac p q a_n\}+\{m_n\cdot \frac 1 q\}\}$ i think...
In last case: when you wrote
[quote]
We have also  \[ b_{n+1}=[\frac p q b_n+\frac p q a_n]=\frac p q (b_n-m)+[\frac p q a_n] +[\{\frac p q a_n\}+\{\frac p q m\} ] \]
[/quote]
you mean
 \[ b_{n+1}=[\frac p q b_n+\frac p q a_n]=[\frac p q (b_n-m)+\frac p q m+\{\frac p q a_n\} +[\frac p q a_n]] = \]
\[ = [\frac p q (b_n-m)+\frac p q m+\{\frac p q a_n\}] +[\frac p q a_n] \] and if $\frac p q (b_n-m) \in Z$
\[ =\frac p q (b_n-m)+[\frac p q m+\{\frac p q a_n\}] +[\frac p q a_n] = \frac p q (b_n-m)+ [\frac p q a_n]+[\{\frac p q m\}+\{\frac p q a_n\}]+[\frac p q m] \].  Then must be $[\frac p q m]=0$ and $pm<q$.
If $m_n=(p\cdot b_n) \text{mod q}$ then $pm<q$, but why then $\frac p q (b_n-m) \in Z$. :?:
Maybe you mean $\frac{pb_n-m}{q}$ :?:
So in any case i can't understand the begin of your solution...
Sorry... Can you explain? :stink:

I've edited it but now I cannot get to any contradiction. I think that from (4) we can get to contradiction using a number theory argument. But I don't know it now. Sorry for mistakes. :blush:

All right. Please write if you will find a contradiction.

[quote="xirti"]I've edited it but now I cannot get to any contradiction. [/quote]
Actually, [b]xirti[/b] has done all the hard work! What remains to check is that the sequence defined by $a_{n+1}=T(a_n): =\left\{\frac pq a_n+ \frac mq\right\}$ where $m$ is a fixed integer, cannot have a limit $\ell$ unless $a_n=\ell$ for all sufficiently large $n$. This follows immediately from the observation that any fixed point $\ell$ of $T$ is a repelling point, i.e., if $|a-\ell|$ is small but not exactly $0$, then $|T(a)-\ell|$ is greater than $|a-\ell|$. On the other hand, we cannot have $a_n=\left\{\left(\frac pq\right)^n\right\}=\ell$ for all sufficiently large $n$ because the denominator of $a_n$ is $q^n$, which changes with $n$, so all $a_n$ are different.

Good observation. Thank you for finishing my work! I dind't thought of this argument!

Great thanks xirty and fedja !!!  :)  :)


[quote="tang zy"]prove that: for every positive real numbers $ x,y,z$,we have $ \sum \frac {\sqrt {(x \plus{} y)(x \plus{} z)}}{2x}\geq \sum \sqrt {\frac {2x}{y \plus{} z}}$[/quote]
Sorry,this is easy:
$ \sum \frac {\sqrt {(x \plus{} y)(x \plus{} z)}}{x}\geq \sum (\sqrt{\frac{y}{x}}\plus{}\sqrt{\frac{z}{x}})\equal{}\sum (\sqrt{\frac{x}{z}}\plus{}\sqrt{\frac{x}{y}})\geq 2\sum \sqrt{\frac{2x}{y\plus{}z}}$
but the following is harder:
prove that: for every positive real numbers $ x,y,z$,we have $ \sum \frac {\sqrt {(x \plus{} y)(x \plus{} z)}}{2x}\geq \sum \sqrt {\frac {y\plus{}z}{2x}}$
Can anyone prove it or disprove it?

[quote="tang zy"]
but the following is harder:
prove that: for every positive real numbers $ x,y,z$,we have $ \sum \frac {\sqrt {(x \plus{} y)(x \plus{} z)}}{2x}\geq \sum \sqrt {\frac {y \plus{} z}{2x}}$
Can anyone prove it or disprove it?[/quote]
I think your inequality is true.
Let $ yz \equal{} a,$ $ xz \equal{} b$ and $ xy \equal{} c.$
Hence, $ \sum_{cyc} \frac {\sqrt {(x \plus{} y)(x \plus{} z)}}{2x}\geq \sum_{cyc} \sqrt {\frac {y \plus{} z}{2x}}\Leftrightarrow\sum_{cyc}\sqrt {a(a \plus{} b)(a \plus{} c)}\geq\sum_{cyc}a\sqrt {2(b \plus{} c)},$
which follows from known $ \sum_{cyc}\sqrt {a(a \plus{} b)(a \plus{} c)}\geq2\sqrt {(a \plus{} b \plus{} c)(ab \plus{} ac \plus{} bc)}$ and
from $ \sqrt {2(a \plus{} b \plus{} c)(ab \plus{} ac \plus{} bc)}\geq\sum_{cyc}a\sqrt {b \plus{} c},$ which is C-S. :wink:

[quote="MathUniverse"][quote="arqady"]
which follows from known $ \sum_{cyc}\sqrt {a(a \plus{} b)(a \plus{} c)}\geq2\sqrt {(a \plus{} b \plus{} c)(ab \plus{} ac \plus{} bc)}$ [/quote]

That is well known?  :)[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=221359
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=280805


okay so the problem is:

Find the number of points on the curve $\displaystyle\frac{1}{x} + \displaystyle\frac{1}{y} = \displaystyle\frac{1}{2004}$  whose coordinates are positive integers.

I'll finish it for you :D

[tex]\displaystyle (2004-x)(2004-y)=2004^2[/tex]
[tex]\displaystyle 2004^2=2^4\times3^2\times167^2[/tex]
So [tex]2004^2[/tex] has [tex]45[/tex] factors.
Each set of factors produces a distinct [tex](x, y)[/tex], so there are [tex]45[/tex] points on the curve with positive integer coordinates.

And just for the record, I missed it too.  :blush:


inequalities
rearrangement inequality
inequalities unsolved

let a,b,c are non-negative such that $ a^2\plus{}b^2\plus{}c^2\equal{}3$.prove that:
$ a^2b^2\plus{}b^2c^2\plus{}c^2a^2\plus{}2(a^3b^3\plus{}b^3c^3\plus{}c^3a^3) \leq 9$

[quote="tuantam1lan"]let a,b,c are non-negative such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$.prove that:
$ a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2 \plus{} 2(a^3b^3 \plus{} b^3c^3 \plus{} c^3a^3) \leq 9$[/quote]
We homogenise and we have
$ \frac {(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2)(a^2 \plus{} b^2 \plus{} c^2)}{3} \plus{} 2(a^3b^3 \plus{} b^3c^3 \plus{} c^3a^3) \leq \frac {(a^2 \plus{} b^2 \plus{} c^2)^3}{3}$
Which is:
$ \sum_{sym}a^4b^2 \plus{} 3a^2b^2c^2 \plus{} 6 \sum a^3b^3 \leq a^6 \plus{} b^6 \plus{} c^6 \plus{} 3\sum_{sym}a^4b^2 \plus{} 6a^2b^2c^2 \Leftrightarrow 6 \sum a^3b^3\leq a^6 \plus{} b^6 \plus{} c^6 \plus{} 3a^2b^2c^2 \plus{} 2\sum_{sym}a^4b^2$
But from Schur we have that:
$ a^6 \plus{} b^6 \plus{} c^6 \plus{} 3a^2b^2c^2 \geq \sum_{sym}a^4b^2$
So we finaly have to prove that:
$ \sum_{sym}a^4b^2 \geq 2a^3b^3$ which is muirihead :wink:

since $ a^2\plus{}b^2\plus{}c^2$=3 therefore $ ab\plus{}bc\plus{}ca$ $ \le3$
squaring onboth sides ,the inequality can easily be proved by rearrangement inequality

Sophie Germain, you are wrong.
What you say is equivalent to :
$ \sum {a^3b^3} \le abc(\sum a)$.
EDIT:{sorry that I forgot $ \sum a^2\equal{}3$.But even with that what sophie germain said is wrong.
But this is not true. Check for $ a \equal{} \sqrt{2},b \equal{} \sqrt{2},c \equal{}\sqrt{\frac{1}{2}}$. We get that $ \frac{17}{8}\equal{} 2.125 < 2$ which is wrong.

[quote="tuantam1lan"][quote="Dimitris X"][quote="tuantam1lan"]let a,b,c are non-negative such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$.prove that:
$ a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2 \plus{} 2(a^3b^3 \plus{} b^3c^3 \plus{} c^3a^3) \leq 9$[/quote]
We homogenise and we have
$ \frac {(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2)(a^2 \plus{} b^2 \plus{} c^2)}{3} \plus{} 2(a^3b^3 \plus{} b^3c^3 \plus{} c^3a^3) \leq \frac {(a^3 \plus{} b^3 \plus{} c^3)^3}{3}$
Which is:
$ \sum_{sym}a^4b^2 \plus{} 3a^2b^2c^2 \plus{} 6 \sum a^3b^3 \leq a^6 \plus{} b^6 \plus{} c^6 \plus{} 3\sum_{sym}a^4b^2 \plus{} 6a^2b^2c^2 \Leftrightarrow 6 \sum a^3b^3\leq a^6 \plus{} b^6 \plus{} c^6 \plus{} 3a^2b^2c^2 \plus{} 2\sum_{sym}a^4b^2$
But from Schur we have that:
$ a^6 \plus{} b^6 \plus{} c^6 \plus{} 3a^2b^2c^2 \geq \sum_{sym}a^4b^2$
So we finaly have to prove that:
$ \sum_{sym}a^4b^2 \geq 2a^3b^3$ which is muirihead :wink:[/quote]
it is shur inequality
 we have:$ a^2b^2(1 \plus{} 2ab) \leq a^2b^2(1 \plus{} a^2 \plus{} b^2) \equal{} a^2b^2(4 \minus{} c^2)$
we must prove that:$ 4(a^2b^2 \plus{} c^2b^2 \plus{} c^2a^2) \minus{} 3a^2b^2c^2 \leq 9$
 $ \leftrightarrow 4(a^2b^2 \plus{} c^2b^2 \plus{} c^2a^2)(a^2 \plus{} b^2 \plus{} c^2) \minus{} 9a^2b^2c^2 \leq (a^2 \plus{} b^2 \plus{} c^2)^3$[/quote]
I made a careless mistake in the first line.The $ RHS$ wasn't $ \frac {(a^3 \plus{} b^3 \plus{} c^3)^2}{3}$ but $ \frac {(a^2 \plus{} b^2 \plus{} c^2)^2}{3}$.I edited
I think my solution is right and pretty easy :)


:arrow: let $ a,b,c > 0$, prove that:

$ \frac {a^2}{a^2 \plus{} ab \plus{} b^2} \plus{} \frac {b^2}{c^2 \plus{} bc \plus{} b^2 } \plus{} \frac {c^2}{c^2 \plus{} ca \plus{} a^2}\geq1$ :?:


Let there be $ x , y , z > 0$ . Prove that :

$ \arctan(x) \plus{} \arctan(y) \plus{}  \arctan(z) < \pi$ if and only iff $ xyz < x \plus{} y \plus{} z$ .

Edited


geometry
3D geometry
number theory open
number theory

Find all positive integer solutions of following equations:

1) $ a^2\plus{}b^2\equal{}2008$

2) $ a^3\plus{}b^3\plus{}c^3\equal{}2007$

3) $ a^3\plus{}b^3\plus{}c^3\equal{}2008$

4) $ a^4\plus{}b^4\plus{}c^4\plus{}d^4\equal{}2007$

5) $ a^4\plus{}b^4\plus{}c^4\plus{}d^4\equal{}2008$

Sorry if it's so easy :(

There are only a few cases to try,
plus for example if you consider mod 4 there are even fewer cases.
WLOG $ a\ge b \ge c \ge d$
Notice that $ a^2 < 2008 \Rightarrow a<45$
$ a^3 < 2008 \Rightarrow a<13$
$ a^4 < 2008 \Rightarrow a<7$

But there are still many cases to try, please give another solutions

1) WLOG $ a\ge b$
Notice that $ a^2 < 2008 \Rightarrow a \ge 44$
$ 2 a^2 \ge a^2 \plus{} b^2 \Rightarrow a^2 \ge 1004 \Rightarrow a \ge 32$
No solution for the cases of $ a \equal{} 32,33,\cdots, 44$

2) and 3) $ 9 \le a \le 12$
$ a\equal{}9$
$ b^3\plus{}c^3\equal{}1278 or 1277$
$ b \le a\equal{}9$ and $ 2b^3 \ge 1278 or 1277$ then $ b\equal{}9$
then $ c^3\equal{}549 or 548$ no solution.

and so on ...

1) Is trivial if you would stop attacking it by casework  :wink:

3) and 4)
We know that for any 'a' belonging to integers,
                     a^4=1 or 0 mod 16
Therefore,
         0<= a^4 + b^4 + c^4 + d^4 <= 4(mod 16)
       But,2007=7(mod 16)and 2008=8(mod 16)
        Thus the given eqns have no soln  :)

For 1, first use mod 4 to get $ a\equal{}2x,b\equal{}2y$, so that $ x^2\plus{}y^2\equal{}502$. Now $ x,y$ have same parity, so $ p\equal{}\frac{x\plus{}y}{2},q\equal{}\frac{x\minus{}y}{2}\Rightarrow p^2\plus{}q^2\equal{}251$. But this has no solution! (It fails modulo 4.)

For 2, use mod 7. Cubes are $ 0,\pm 1$ mod 7, while 2007 is -2. The only case is where two of the cubes are -1 and the third is 0. So 7 divides one of $ a,b,c$ (WLOG let it be $ c$). Then $ c\equal{}7$, because $ 14^3>2007$. So we have
\[ (a\plus{}b)(a^2\minus{}ab\plus{}b^2)\equal{}1664\equal{}2^7\cdot 13\]
Note that $ 4(a^2\minus{}ab\plus{}b^2)\equal{}(a\plus{}b)^2\plus{}3(a\minus{}b)^2\geq (a\plus{}b)^2$ so $ (a\plus{}b)^3\leq 2^9\cdot 13\Rightarrow a\plus{}b\leq 2^3\sqrt[3]{13}<2^5$.

This means $ a^2\minus{}ab\plus{}b^2$ is divisible by $ 2^3$. Now $ a^2\minus{}ab\plus{}b^2$ can be even only if $ a,b$ even, and in this case, 2 can be factored out of each number.

We find that $ a,b$ both divisible by 4. Let $ a\equal{}4x,b\equal{}4y$ so that $ x^3\plus{}y^3\equal{}2\cdot 13$. But this clearly has no solution:
\[ x^3\plus{}y^3<3^3\Rightarrow x,y\leq 2\Rightarrow x^2\plus{}y^2\leq 16\]


geometry
perimeter
trigonometry
geometry solved

Let $a, b, c$ be sides of a triangle whose perimeter does not exceed $2 \cdot \pi.$, Prove that $\sin a, \sin b, \sin c$ are sides of a triangle.

http://www.mathlinks.ro/Forum/viewtopic.php?t=5606 .

  darij


function
LaTeX
algebra
rational function
superior algebra
superior algebra unsolved

Let k be a Field, K = k(x) the rational function field, a in Aut_k(K) given by a(x) = 1/(1-x) and G =<a>. Show that |G| = 3 and determine K^G.

Now let a(x)=x+1 and assume that char(k)=0.  Show that G is infinite and determine K^G.

When you reply, can you explain how to put LaTeX equations into my post?  I know how to write in TeX, but I don't know how to include my equations into the forum.  Thanks!

Just put dollar signs around your math mode $ \text{\LaTeX}$.

For the first problem, just compute $ a^3$: $ a^2(x)=a(\frac1{1-x})=\frac1{1-\frac1{1-x}}=\frac{x-1}{x}$, and $ a^3(x)=a(\frac{x-1}{x})=\frac{\frac1{1-x}-1}{\frac1{1-x}}=x$.
Anything that fixes $ x$ fixes everything.

What's fixed by the automorphism? Any expression that's symmetric in $ x,\frac1{1-x}$, and $ \frac{x-1}{x}$ is. That includes the sum $ \frac{-x^3+3x-1}{x(1-x)}$; it should generate the fixed field. The full quotient field will have degree 3 over the fixed field.


What conditions do we need for $ a, b, c$ so that the following inequality holds:
\[ a^3 \plus{} b^3 \plus{} c^3 \plus{} 6abc \geq 2(a^2b \plus{} b^2c \plus{} c^2a) \plus{} ab^2 \plus{} bc^2 \plus{} ca^2
\]
For example, you may say, $ a,b,c$ should be sides of triangle; or you may say $ a,b,c$ should be positive real numbers between 1 and 2.
or, $ a, b , c$ should be diagonals of hexagon  :P

anyways, i am interested in this inequality very much. i need your help! Thanks!

It's already posted by darij grinberg that
If $ a, b, c$ be ther lengths of a triangle.
$ a^{3}\plus{}b^{3}\plus{}c^{3}\plus{}3abc\minus{}2ab^{2}\minus{}2bc^{2}\minus{}2ca^{2} \ge 0$
and
$ 3a^{2}b\plus{}3b^{2}c\plus{}3c^{2}a\minus{}3abc\minus{}2ab^{2}\minus{}2bc^{2}\minus{}2c^{2}a \ge 0$

and well known equality
$ 2(ab^{2}\plus{}bc^{2}\plus{}ca^{2})\equal{}a^{2}b\plus{}b^{2}c\plus{}c^{2}a\plus{}3abc$ Holds.
So your problem is done.


Find the area of a circle described by the equation: $ 4x^2\plus{}4y^2\plus{}24x\minus{}20y\minus{}3\equal{}0$

A. $ 4\pi$
B. $ 12\pi$
C. $ 16\pi$
D. $ 18.25\pi$
E. None of the Above

[hide]We complete the square to get the equation $ (x\plus{}3)^2\plus{}(y\minus{}\frac{5}{2})^2\equal{}16$. Thus, the area is $ 16\pi$, or C.[/hide]

Hint: Try rewriting the circle equation in the form (x-h)^2+(y-k)^2=r^2, and find r^2. Then multiply it by pi, and you get the circle's area.

I know this is high school, but if you know calculus, just set up a double integral.  It's really tedious and really annoying but if you do it right, you'll feel proud of yourself.

Some people don't waste extra time doing something more complicated just to feel proud of themselves.  :wink:


can somebody solve step by step this problem:

10x^2+70x-180

$ 10x^{2}\plus{}70x\minus{}180 \equal{} 10(x^{2}\plus{}7x\minus{}18)$

Thus, all you need to find is two numbers, say, a and b, such that a+b=7 and (a)(b) = -18. Once you are done finding them, you can conclude that

$ 10x^{2}\plus{}70x\minus{}180 \equal{} 10(x^{2}\plus{}7x\minus{}18) \equal{} 10(x\plus{}a)(x\plus{}b).$

solve the quadratic for 0? If so,

$ y \equal{} 10x^{2}\plus{}70x\minus{}180$
$ y \equal{} x^{2}\plus{}7x\minus{}18$
$ y \equal{} (x\minus{}2)(x\plus{}9)$
:. $ 0 \equal{} (x\minus{}2)(x\plus{}9)$

hence solutions are:  $ \boxed{x: (2,\minus{}9)}$

Or you could use the Quadratic Formula to get the answer.

[quote="abacadaea"]if you are dealing with a leading coefficient of one, this is easier.[/quote]

Yeah.. True..

Guys it looks like people have already solved it. We dont need any further posts repeating the same thing.

Mods please lock this.

we were just discossing easier sulotions   :?:


limit
integration
calculus
calculus computations

some kind of proof would be greatly appreciated.  :read:

Let L=(lim n->infinity)(sigma)(n(n-1)(n-2)....(n-k+1))/(n.n.n.....ktimes)

=(lim n->infinity)(sigma)(1-1/n)(1-2/n).....(1-(k-1)/n)

=1.

The question seems to be erroneous.

[hide="Possible Idea"]
$ n! \approx \sqrt{2\pi n}\left(\frac{n}{e}\right)^n$, can we use this approximation?[/hide]

For any fixed $ k$, the $ k$th term increases to $ 1$ as $ n\to\infty$. The limit of the sum is infinite.

Can you say how quickly?  (By mathematica, it appears to grow like $ \sqrt{n}$, but I wasn't able to come up with an argument.)

Yes, it does grow like $ \sqrt{n}$: $ \sqrt{\frac{\pi n}{2}}$ to be specific.

The $ k$th term is $ \exp\left(\sum_{j\equal{}1}^k\ln(1\minus{}\frac jn)\right)\approx \exp\left(\minus{}\frac{k(k\plus{}1)}{2n}\right)$ for $ k<<n$. The sum of these terms is approximately $ n\int_0^1 e^{\minus{}nx^2/2}\,dx$; as $ n\to\infty$, this is $ \sqrt{n}\int_0^{\infty}e^{\minus{}x^2/2}\,dx\equal{}\sqrt{\frac{\pi n}{2}}$.


geometry
incenter
circumcircle
geometry unsolved

In a triangle ABC, $ \hat{A}$ = $ 20\text{\textdegree}$ and $ \hat{B}$ = $ 110\text{\textdegree}$. If $ I$ is the incenter (center of the circle inscribed) and $ O$ is the circumcenter (center of the circle) of the triangle ABC, which measure of the angle $ I\hat{AO}$ ?

Oi luis eduardo

Não deu prá entender. Se quiser, "poste" em português  que a gente tenta verter...

I'm sorry, in the question I forgot to put the angle question. See the question again please.

That's ok ;)

Portuguese question:

Em um triângulo ABC,  A = 20 graus e  B = 110 graus. Se I é o incentro (centro da circunferência inscrita) e O o circuncentro (centro da circunferência circunscrita) do triângulo ABC, qual a medida do ângulo I^AO ?

I think the answer is $ 30\text{\textdegree}$. That's easy...

O is in the outside of triangle ABC.

Because ABC is an obtuse triangle.

So, $ I\hat{AC}$ = 1 0

$ C\hat{AO}$ = 20

So, $ I\hat{AO}$ = 30


Prove that every infinite sequence $S$ of distinct positive integers contains either an infinite subsequence such that for every pair of terms, neither term ever divides the other, or an infinite subsequence such that in every pair of terms, one always divides the other.

We call two terms $ a_i$ and $ a_j$ friends if $ a_i|a_j$ or $ a_j|a_i$ and enemies otherwise.We call a sequence "good" if any two terms are friends.

Assume there is no good infinite sequence.

Let $ \displaystyle a_1=a{}_1{}_1, a{}_1{}_2,\dots a{}_1{}_n$  be the longest good sequence containing $ a_1$, where $ 1_1, 1_2, \dots 1_n$ are indices.This means that for any $ \displaystyle i$ distinct from $ \displaystyle \{1_1, 1_2, \dots 1_n\}$, there exists $ j\leq n$ such that $ a_i$ and $ a{}_1{}_j$ are enemies.

It's clear that there exists a number $ a{}_1{}_k$ with an infinite number of enemies. Denote $ a{}_1{}_k=b_1.$Therefore there exists an infinite set $ I_1$ of indices such that for any $ i\in I_1$, $ a_i$ and $ b_1$ are enemies.

Assume that we have constructed the numbers $ b_1, b_2\dots b_n$ such that any two are enemies and there exists an infinite set $ I_n$ of indices such that for any $ b_i$ and for any $ j\in I_n$, $ b_j$ and $ a_i$ are enemies.

Pick a number $ x$ from $ I_n$ such that $ a_x$ is distinct from $ b_i$, for any $ i\leq n$, and let $ a_x=a{}_n{}_1, a{}_n{}_2, \dots a{}_n{}_m$ be the longest good sequence containing $ a_x$, with indices $ a{}_n{}_i\in I_n$, for any $ i\leq m$. Therefore any element $ y$ of $ I_n$, which is not from the set $ \{n_1, n_2\dots n_m\}$, has the property that $ a_y$ is an enemy with some $ a{}_n{}_j$.

Since the set $ I_n -\{n_1, n_2\dots n_m\}$ is infinite, there exists an indice $ n_k$ which is an enemy with an infinite number of elements with indices from $ I_n$.. Let $ a{}_n{}_k=b_{n+1}$ and let $ I_{n+1}\subset I_n$ be the set of indices with which $ b_{n+1}$ is an enemy.

Therefore we have constructed the number $ b_{n+1}$ which is an enemy with $ b_1, b_2\dots b_n$, and the infinite set of indices $ I_{n+1}$ such that for any $ y\in I_{n+1}$, $ a_y$ is an enemy with any $ b_{i}$, $ \forall$ $ i\leq n+1$.

Thus, by induction, we have constructed an infinite subsequence $ \{b_n\}$ of enemies.

The problem is just a nice application of [url=http://mathworld.wolfram.com/DilworthsLemma.html]Dillworth theorem[/url],which is similar with [url=http://www.mathlinks.ro/viewtopic.php?p=198301#198301]Romania 2005 TST Day 1 problem 2[/url]


in right triangle ABC with right angle C, angle B=41 degrees. Square PQRS is inscribed in the right triangle. let h be the altitutude from C to AB. if , $ \frac {1}{h} \plus{} \frac {1}{AB} \equal{} \frac {2}{3}$, what is the length of a side of the square?

and a friend quizzed me on this  :D

nvm i got it :)

[hide="half of a solution maybe?"]
DC is the perpendicular from C to AB.
Draw the median CE of the triangle.  Call the length of the median r.  It's easy to see that $ \frac {r}{h} \equal{} \cos 8$, or $ \frac {AB}{2h} \equal{} \cos 8$ or $ AB \equal{} 2h\cos 8$  Then, using the equation, convert it to $ AB \plus{} h \equal{} \frac {2AB\cdot h}{3}$
Divide by h on both sides. $ \cos 8 \plus{} 1 \equal{} \frac {2AB}{3}$ or $ AB \equal{} \frac {3 \plus{} 3\cos 8}{2}$?

[/hide]
[hide="only correct if the first half is right"]
The side of the square is s.
Similar triangles to get $ s(\frac{1}{\tan 41}\plus{}\tan 41\plus{}1)\equal{}AB\equal{}\frac{3\plus{}3\cos 8}{2}$

$ s\equal{}\frac{3\plus{}3\cos 8}{\frac{2}{\tan 41}\plus{}2\tan 41\plus{}2}$

Too lazy to calculate.
[/hide]
EDIT:edited in the second half


[quote="returnofkiladi"]Please help me out with these problems (answer & explain, plz):
2) How many full seven-day weeks are there in seven consecutive years. Assume the first year in the first day of the week.
6) Two complementary angles, A and B, have measures in the ratio of 7 to 23, respectively. What is the ratio of the measure of the complement of angle A to the measure of the complement of angle B?
17) What is the least natural number that can be added to 40,305 to create a palindrome (plz explain).[/quote]
2. [hide]Each year has $ 52$ full weeks, so in $ 7$ consecutive years, there are $ 52$ $ *$ $ 7$ $ \equal{}$ $ 364$ full seven day weeks.[/hide]
6. [hide]Let $ A$ be $ 7x$ and $ B$ be $ 23x$. Since they're complementary, they must sum to $ 90$ degrees. $ 30x$ $ \equal{}$ $ 90$, so $ x$ $ \equal{}$ $ 3$. Angle $ A$ measures $ 21$ degrees and $ B$ measures $ 69$ degrees. Ratio is $ 69$ $ /$ $ 21$ $ \equal{}$ $ 23$ $ /$ $ 7$.[/hide]
17. [hide]With $ 3$ as the hundreds digit, $ 40,304$ is the least palindrome, so the next least one is $ 40,404$. $ 40,404$ $ \minus{}$ $ 40,305$ $ \equal{}$ $ 99$. $ 99$ is the least natural number you can add to $ 40,305$ to create a palindrome.[/hide]

2. [hide="actually,"]Since there are 365 days in a year and 7 in a week, we multiply 365 days by 7 years, then divide by 7weeks to get $ \boxed{365}$ full weeks[/hide]

Gauss, your solution doesn't take in the case if the extra days other than the 52 weeks make another week, which it does.

6. [hide]The complement of angle A is B and the complement of B is A. Since A/B=7/23 and we want B/A, we raise 7/23 to the -1 power to get $ \boxed{\frac{23}{7}}$[/hide]

17. [hide]Since the next palindrome  has a ten thousand digit of 4, the units digit must be 4. The next hundreds digit is 4, so the next palindrome is 40404. Thus, 40404-40305=$ \boxed{99}$[/hide]

Wait are you mentioning leap years for extra days?

No. Imagine instead of a period of 7 years, a period of at least $ 7\cdot365$ (without leap years) days. Since every 7 days is a week (lol, why am I saying this), we must have $ 7\cdot365/7\equal{}365$ full weeks. The leap years are negligible, since there are at most 2 extra days in the period, which wouldn't result in more weeks.

Thanks guys.
I rlly appreciate the help.
The reason why I even posted number 2, was because I got 364, also. I was just wondering how the answer was 365.


It's my old problem  :wink:
[quote="nguoivn"]Given $ a, b, c > 0$. Prove that: $ \frac {a}{\sqrt {b^2 \plus{} ab \plus{} 9c^2}} \plus{} \frac {b}{\sqrt {c^2 \plus{} bc \plus{} 9a^2}} \plus{} \frac {c}{\sqrt {a^2 \plus{} ca \plus{} 9b^2}} \ge \frac {3}{\sqrt {11}}$
[/quote]
See also here: http://www.mathlinks.ro/viewtopic.php?t=299970


What are the chances of getting a job in a good hedge fund if you are a top 50 Divison 1 athlete with a double major in econ and math? Do you think hedge funds would consider being an athlete as an advantage or they would prefer someone amazing in math? Thanks.

[quote="rrusczyk"]Depends on the position.  Certainly proven excellence in anything is a plus.  But if you're looking for a quantitative modelling position, being a great athlete will be much less helpful in an interview than if you're looking for a trading, marketing, or sales position, where your interviewers will be more likely to be ex-athletes and more likely to see higher value in someone who has developed those skills.[/quote]

why does trading, marketing or sales position tend to be more popular with ex-athletes?

[quote="rrusczyk"]I think it has more to do with there not being a huge intersection between 'top notch athletes' and 'top notch quants'.  To be a top notch anything requires a lot of time.  To do so in both athletics and math/CS is very hard.  It's absolutely necessary to do the latter to be a top quant.  You can become a very good marketer/trader/etc without having developed very strong quant skills.[/quote]

good point. i think i'm somewhere in between then.

come to think of it, i can't think of anyone thats in that intersection. i know a few quants and i've been around many athletes. but none that i can think of is excellent in both.

This is perhaps the most perfect example of why you need more than one version of your CV.

I counsel you to create >2 versions, each with a different emphasis, from full on sport, to barely mentioning it.

My reason is that you can't know the utility function of those who look at your CV, indeed to get a job it will often be read by more than one person. In that uncertain state, you should not rely upon one play for all jobs. In the absence of more precise into about a given employer you should choose randomly from your portfolio of CV versions.

Although you've said absolutely nothing that leads me to doubt your integrity, the advice I give has led some people to think I am advocating that "multiple versions" means adding things that aren't as true as they ought to be. That is absolutely not the case. CVs are checked, and I have any number of anecdotes about people who have gilded the rose being blown up at interview.


Prove that any polynomial can be expressed as the difference of two monotonically increasing polynomials.

we take $ f,g: \mathbb{R}[X]\to \mathbb{R}[X]$ such that:
$ \forall a\in \mathbb{R}: f(a)\equal{}a,g(a)\equal{}0$
for $ n\in\mathbb{N}^*$
for $ a\ge 0: \ f(aX^{2n})\equal{}\frac{aX^{2n\plus{}1}}{2n\plus{}1}\plus{}\frac{2aX^{2n}}{2n}\plus{}\frac{aX^{2n\minus{}1}}{2n\minus{}1}$
and we have $ (\frac{aX^{2n\plus{}1}}{2n\plus{}1}\plus{}\frac{2aX^{2n}}{2n}\plus{}\frac{aX^{2n\minus{}1}}{2n\minus{}1})'\equal{}aX^{2(n\minus{}1)}(X\plus{}1)^2\ge 0$
then $ f(aX^{2n})$ is increasing.
and $ g(aX^{2n})\equal{}\frac{aX^{2n\plus{}1}}{2n\plus{}1}\plus{}\frac{aX^{2n\minus{}1}}{2n\minus{}1}$ is increasiong.
if $ a<0$ $ f(aX^{2n})\equal{}g(\minus{}aX^{2n})$ and $ g(aX^{2n})\equal{}f(\minus{}aX^{2n})$
and for $ a\ge 0$ $ f(aX^{2n\plus{}1})\equal{}aX^{2n\plus{}1},g(aX^{2n\plus{}1})\equal{}0$
for $ a<0$ $ f(aX^{2n\plus{}1})\equal{}0,g(aX^{2n\plus{}1})\equal{}\minus{}aX^{2n\plus{}1}$

so $ \forall (a,n)\in\mathbb{R}\times \mathbb{N}: aX^n\equal{}f(aX^n)\minus{}g(aX^n)$ and $ f(aX^n),g(aX^n)$ are increasiong.

$ \boxed{P(X)\equal{}\sum_{i\equal{}0}^{n}a_iX^i\equal{}\sum_{i\equal{}0}^{n}f(a_iX^i)\minus{}\sum_{i\equal{}0}^{n}g(a_iX^i)}$

[quote="ali666"][quote="aviateurpilot"]$ P(x) \equal{} \sum_{i \equal{} 0}^{n}a_ix\i \equal{} \sum_{a_i > 0}a_ix^i \minus{} \sum_{a_i\le 0}( \minus{} a_i)x^i$[/quote]
if a polynomial have  nonnegative coefficient,it is not necessarily increasing!for example $ x^2$[/quote]
yes  :blush: ,thnkx
but i ideted my solution


Here's a nice question: identify products A and B in the following reactions,

1) Epichlorohydrine + CH3OH/H2SO4 ------> A

2) A + NaHO/Cl2 ------> B

[hide="answer"]
Ring opening takes place in the first. More stable carbocation in this case is the primary. :maybe:
$ \ce{A \minus{} Cl \minus{} CH2 \minus{} CH(OH) \minus{} CH2 \minus{} OCH3}$
And then the haloform reaction is it? And the ether is not disturbed I think. :maybe:
so $ \ce{B \minus{} CH3O \minus{} CH2 \minus{} COONa}$
[/hide]

Perfect Chemrock, that is the correct answer.


Convert the angle measures to degrees, minutes, and seconds 42.96''


I am using MathLinks, and the "go to page" feature in the memberlist is not working. Nothing happens when I click on the button?

It doesn't work for me too. I just change the URL which changes the page. If you increase the number at the end by 50, it goes to the next page. For example, if you have 150 at the end (page 4) and change it to 550 (page 12), you will go forward 8 pages.

Yes, it's working.
(For the record: Click where it says "Page _ of _", then type in the page number. Also applies to forums and the like.)


geometry
3D geometry
polygon
cube
Intersection
IMO Shortlist
IMO Longlist

Which regular polygon can be obtained (and how) by cutting a cube with a plane ?

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


Divide both sides by $ 81^x$ (which is always nonzero) and rearrange to get $ 3\left(\frac49\right)^{2x} \plus{} 37\left(\frac49\right)^x \minus{} 26 \equal{} \left(3\left(\frac49\right)^x \minus{} 2\right)\left(\left(\frac49\right)^x \plus{} 13\right) \equal{} 0$
$ \implies\left(\frac49\right)^x \equal{} \left\{\frac23, \minus{} 13\right\},\left(\frac49\right)^x\ge0\implies\left(\frac49\right)^x \equal{} \frac23\implies x \equal{} \boxed{\frac12}$ (there's only one solution since the LHS is monotonically decreasing).


trigonometry
calculus
AMC 10
AMC 10 A
geometry
trig identities
Law of Sines

I know that it isn't necessary, but by the looks of it, it's pretty useful.

For example, number 25 on AMC 10 A 2005 was easily solved with trigonmetry.

http://www.artofproblemsolving.com/Wiki/index.php/2005_AMC_10A_Problems/Problem_25

What trigonometric formulas, identities, etc, should I know?  The only trig I know is the unit circle, a couple identities, like double and half angle, and really really basic triangles.  (Like find the sin of a given angle theta of a right triangle with legs 3 and 4.)

You can look over chapter 2 and chapter 3 of volume 2 for a lesson on trigonometry. It basically goes over all the important things needed for AMC/AIME, etc.

Shouldn't there be an alternate non-trig solution?

I think there's one with similarity, but I'm not sure.

[quote="athunder"]I would recommend knowing law of cosines and law of sines as well. Other then that AMC 10 will not have anything else dealing with trigs.[/quote]

But for law of sines and law of cosines, wouldn't you need a calculator to find the sines and cosines of some of the angles?

yaofan, in contests where calculators aren't allowed, like the AMC 10, the problems do not require you to find the sine or cosine of a complicated angle, most of the times, it is either you have to use trigonometric identities to just solve for, say for example $ \tan \angle BAC,$ without actually knowing what $ \angle BAC$ is. Or you are given angles that are easy to deal with (like, $ \sin 45, \sin 30,$ etc.). Just draw a unit circle to help with the signs.

Another trick for the signs is "ASTC" or All Students Take Calculus. I tells you what's positive in which quadrant
quadrant 1 -A-all (all are positive)
quadrant 2 -S-students (sins and cosecants are positive)
quadrant 3 -T-take (tangents and cotangents are positive)
quadrant 4 -C-calculus (cosines and secants are positive)

I've been using the "ASTC" rule for a while, and it works quite well.   :D

I'm pretty lacking in seeing and utilizing similarity though.  How can I practice this?


Go to your profile, choose "Preferences", choose "Posting a message", and change the "Always attach my signature" button to Yes.

[quote="AIME15"]Is there a way to automatically attach your signature to a post?  Becuase I forget to click "attach signature" about 90% of the time :([/quote]
Also, if you didn't attach the signature in the past, the signature won't appear on those posts. (which is very sad)

[quote="AIME15"]Is there a way to automatically attach your signature to a post?  Becuase I forget to click "attach signature" about 90% of the time :([/quote]

Hehe, I would have died laughing if you would have forgotten you sig on that post.... XD


algebra
polynomial
calculus
integration
algebra unsolved

[size=150]Given the polynomial $f(x)=x^n+a_{1}x^{n-1}+a_{2}x^{n-2}+ \cdots +a_{n-1}x+a_n$ with integral coefficients $a_1 ,a_2 , \cdots , a_n$; and given also that there exist four distinct integers $a,b,c$ and $d$ such that $f(a)=f(b)=f(c)=f(d)=5$, show that there is no integer $k$ such that $f(k)=8$.[/size]

f(x)-5=(x-a)(x-b)(x-c)(x-d).
8=f(k)=(k-a)(k-b)(k-c)(k-d)+5
3=(k-a)(k-b)(k-c)(k-d)
some of the numbers: k-a, k-b, k-c,k-d is multiple of 3
Hence the product of the other three is $\pm 1$
Hence there are two of they equals to 1 or to -1(contradiction)


Consider a circle of radius  $33$ cm. and center  $O$. Let  $AOB$  and  $COD$ be two perpendicular diameters of this circle. Consider the two tangent circles of diameters  $AO$  and  $OB$ respectively. We have now three circles. Construct two circles whose centers are on the segment  $CD$  and are tangent to the other three circles. How many centimeters are in the radius of one of the smallest circle?

[hide] Say the the radius of one of the tangent circles is x (they will have the same radius).  Now, say that the center of the circle that is tangent to Circle O at C is P.  Say the center of the circle with OA as diameter is M.

Now, we know that OM=33/2=16.5.

We also know that OC=16.5.

We have that CP=x.

Thus, OP=16.5-x.

Finally, we have that PM=16.5+x.

Since $\angle MOP=90$, we know that $OP^{2}+OM^{2}=MP^{2}$ by the Pythogorean Theorem.

Thus, we have that $(16.5-x)^{2}+16.5^{2}=(16.5+x)^{2}\Rightarrow 16.5^{2}-33x+x^{2}+16.5^{2}=16.5^{2}+33x+x^{2}\Rightarrow 66x=16.5^{2}\Rightarrow x=\frac{16.5^{2}}{66}=(\frac{16.5}{66})16.5=(\frac{1}{4})16.5=4.125$, which is our answer.  [/hide]


I agree with spell it out 11. The answer is 10,000.


Let : $ u_n\equal{}2n^2\minus{}n\minus{}3$ then compute:

$ \lim_{n \to \plus{} \infty} \frac{\sqrt{a_n}\plus{}\sqrt{a_{4n}}\plus{}\sqrt{a_{4^{2}n}}\plus{}...\sqrt{a_{4^{10}n}}}{\sqrt{a_n}\plus{}\sqrt{a_{2n}}\plus{}\sqrt{a_{2^{2}n}}\plus{}...\sqrt{a_{2^{10}n}}}$

Some one knows where it was proposed?


calculus
integration
trigonometry
logarithms
derivative
function
real analysis

Aply

$\ln(a+u) \, = \, \ln a+\sum_{k=1}^{\infty}(-1)^{k+1}\frac{u^{k}}{ka^{k}}$

convergent for $-a < u \le a$

[quote="Kent Merryfield"]Have you tried differentiating with respect to $\lambda?$[/quote]

Yes, I have.
Let $f(\lambda)=\int_{0}^{x}{\ln(1+\lambda\sin t)dt}$
Then $f'(\lambda)=\int_{0}^{x}{\frac{\sin t}{1+\lambda\sin t}dt}=\int_{0}^{\tan{\frac{x}{2}}}\frac{4u}{(1+u^{2})(1+u^{2}+2\lambda u)}du$
=$\frac{2}{\lambda}\int_{0}^{\tan{\frac{x}{2}}}(\frac1{1+u^{2}}-\frac1{1+u^{2}+2\lambda u})du$
=$\frac{2\arctan\tan\frac{x}{2}}{\lambda}-\frac2{\lambda\sqrt{1-\lambda^{2}}}(\arctan\frac{\tan{\frac{x}{2}}+\lambda}{\sqrt{1-\lambda^{2}}}-\arctan\frac{\lambda}{\sqrt{1-\lambda^{2}}})$
Then how to compute $f(\lambda)$?
Moreover, $\tan\frac{x}{2}$ is not defined for all $x>0$.

[quote="cyqwhu"]But I don't think $\sin^{k}t$ can be easily integrated.[/quote]

Sorry :blush:

In the other hand, I think

$f'(\lambda)\not=\int_{0}^{x}{\frac{\sin t}{1+\lambda\sin t}dt}$ but $f'(\lambda)=\int_{0}^{x}{\frac{\cos t}{1+\lambda\sin t}dt}$

[quote="jmerry"]No. The only derivative is taken with respect to $\lambda$; we never differentiate with respect to $t$.

I don't think you can do it for all $x$; that's equivalent to finding an antiderivative, and this function probably doesn't have an elementary antiderivative. The special case $x=2\pi$ should work, and a few others might as well.[/quote]

Oh! yes. I'm slept today


If $ x$ , $ y$ and $ z$ are positive real numbers and $ xyz\equal{}1$ prove that:
$ \frac{1}{1\plus{}x^{3}}\plus{}\frac{1}{1\plus{}y{3}}\plus{}\frac{1}{1\plus{}z^{3}}\leq \frac{xy\plus{}yz\plus{}zx}{2}$

[quote="delegat"]If $ x$ , $ y$ and $ z$ are positive real numbers and $ xyz \equal{} 1$ prove that:
$ \frac {1}{1 \plus{} x^{3}} \plus{} \frac {1}{1 \plus{} y^{3}} \plus{} \frac {1}{1 \plus{} z^{3}}\leq \frac {xy \plus{} yz \plus{} zx}{2}$[/quote]

I'm not sure, but I think that

$ \frac {1}{1 \plus{} x^{3}} \plus{} \frac {1}{1 \plus{} y^{3}} \plus{} \frac {1}{1 \plus{} z^{3}}\leq \frac {1}{1 \plus{} x^{2}} \plus{} \frac {1}{1 \plus{} y^{2}} \plus{} \frac {1}{1 \plus{} z^{2}}$

(in that case, $ \frac {1}{1 \plus{} x^{2}} \plus{} \frac {1}{1 \plus{} y^{2}} \plus{} \frac {1}{1 \plus{} z^{2}} \leq  \frac {xy \plus{} yz \plus{} zx}{2}$ is easy to prove).

Does anybody know is this true?

[quote="delegat"]If $ x$ , $ y$ and $ z$ are positive real numbers and $ xyz \equal{} 1$ prove that:
$ \frac {1}{1 \plus{} x^{3}} \plus{} \frac {1}{1 \plus{} y^{3}} \plus{} \frac {1}{1 \plus{} z^{3}}\leq \frac {xy \plus{} yz \plus{} zx}{2}$[/quote]
Mixing Variables Method kills it.  :wink:

[quote="delegat"]If $ x$ , $ y$ and $ z$ are positive real numbers and $ xyz \equal{} 1$ prove that:
$ \frac {1}{1 \plus{} x^{3}} \plus{} \frac {1}{1 \plus{} y{3}} \plus{} \frac {1}{1 \plus{} z^{3}}\leq \frac {xy \plus{} yz \plus{} zx}{2}$[/quote]
The stronger one still holds :)
\[ \frac {1}{2} \plus{} \frac {1}{x^2 \plus{} y^2 \plus{} z^2 \plus{} xy \plus{} yz \plus{} zx} \ge \frac {2}{xy \plus{} yz \plus{} zx}
\]
:)

Can someone tell what mixing the variables means?

My soln

By power mean inequlaity we have
$ \dfrac{\sum (x^3 \plus{} 1)^ \minus{} 1} {3} \le \dfrac{3}{x^3 \plus{} y^3 \plus{} z^3} \le \dfrac{1}{2}$
Hence we have to prove that $ \dfrac{\sum xy}{2} \ge \dfrac{3}{2}$  which is just a consequence of the AM GM inequality for the positive reals $ xy,yz,zx$.
Is it elegant? Is there any mistake?

[quote="Aravind Srinivas L"]$ \dfrac{3}{x^3 \plus{} y^3 \plus{} z^3} \le \dfrac{1}{2}$[/quote]
This is not true for $ (x,y,z)\equal{}(1,1,1)$

I forgot a $ \plus{} 3$ there akashnil. U should have known it. :wink:

[quote="Aravind Srinivas L"]
$ \dfrac{\sum (x^3 \plus{} 1)^ \minus{} 1} {3} \le \dfrac{3}{x^3 \plus{} y^3 \plus{} z^3\plus{}3}$[/quote]
I think it's the wrong way because
$ ((x^3\plus{}1)\plus{}(y^3\plus{}1)\plus{}(z^3\plus{}1))(\frac{1}{x^3\plus{}1}\plus{}\frac{1}{y^3\plus{}1}\plus{}\frac{1}{z^3\plus{}1}) \ge 9$ by Cauchy.

[quote="djuro"]I'm not sure, but I think that

$ \frac {1}{1 \plus{} x^{3}} \plus{} \frac {1}{1 \plus{} y^{3}} \plus{} \frac {1}{1 \plus{} z^{3}}\leq \frac {1}{1 \plus{} x^{2}} \plus{} \frac {1}{1 \plus{} y^{2}} \plus{} \frac {1}{1 \plus{} z^{2}}$[/quote]

$ x\equal{}y\equal{}\frac12,\ z\equal{}4$.

[quote="djuro"](in that case, $ \frac {1}{1 \plus{} x^{2}} \plus{} \frac {1}{1 \plus{} y^{2}} \plus{} \frac {1}{1 \plus{} z^{2}} \leq \frac {xy \plus{} yz \plus{} zx}{2}$ is easy to prove).

Does anybody know is this true?[/quote]
Indeed, the RHS is $ \frac1{2x}\plus{}\frac1{2y}\plus{}\frac1{2z}$, and $ \sum\left(\frac1{2x}\minus{}\frac1{1\plus{}x^2}\right)\equal{}\sum{\frac{(1\minus{}x)^2}{2x(1\plus{}x^2)}}\geqslant0$. :wink:


advanced fields
advanced fields unsolved

For n >= 1, let f(n) be the number of 1's needed to write all the integers between 1 and n.  For example, f(11)=4.
A balanced number is a fixed point of f, that is, a value of n such that f(n)=n.  There are 84 balanced numbers.

What if we change the base to base d>=2?  The largest balanced number is 1...10, where there are (d-1) 1s.  Let a(d)
be the number of balanced numbers in base d.  Then the sequence {a(d)} is 2, 4, 8, 4, 21, 5, 45, 49, 84, ... .   What is
the pattern?

If you solve it, would you let me know at erickson@truman.edu.  Thank you.

Martin Erickson

when someone posts this many problems in forums whose topics don't match the problems, it's most definitely gotta be homework!


function
topology
advanced fields
advanced fields unsolved

Let $ X,Y,Z$ be topological spaces and suppose $ f_1,f_2\in\mathrm{Hom}(X,Y)$ and $ g_1,g_2\in\mathrm{Hom}(Y,Z)$. Show that if $ f_1$ and $ g_1$ are homotopic to $ f_2$ and $ g_2$ respectively, then the composite functions $ g_1\circ f_1$ and $ g_2\circ f_2$ are homotopic.

Let $ F$ be a homotopy from $ f_1$ to $ f_2$ and $ G$ be a homotopy from $ g_1$ to $ g_2$. Define $ H: X\times[0,1]\to Z$ by $ H(x,t)\equal{}G(F(x,t),t)$. Then $ H(x,0)\equal{}g_1\circ f_1(x)$ and $ H(x,1)\equal{}g_2\circ f_2(x)$. But how to show that $ H$ is continuous? :?:

$ F$ and $ G$ are continuous by definition of homotopy.
Now let the function $ K: X \times [0,1] \rightarrow X \times [0,1]$ defined as $ K(x,t) \equal{} (F(x,t),t)$

The map K is continuous because each component is. Now we notice that $ H \equal{} G \circ K$ so also H is continuous.

bye :)


[quote="Hong Quy"]with all real numbers , we always have ineq :
max {$ x_1,x_2,...,x_n, \minus{} x_1, \minus{} x_2,..., \minus{} x_n$}$ \geq \frac { |x_1| \plus{} |x_2| \plus{} ... \plus{} |x_n|}{2n \minus{} 1}$[/quote]

That's trivial : $ max \{x_1,...,x_n, \minus{}x_1, ..., \minus{}x_n\} \equal{}$$ max \{|x_1|,|x_2|,...,|x_n|\} \geq \frac { |x_1| \plus{} |x_2| \plus{} ... \plus{} |x_n|}n$


topology
integration
advanced fields
advanced fields unsolved

Let $ \omega$ be a form on degree $ k-1$ on the compact $ k$-dimensional manifold $ M$. Prove that $ \int_{M}d \omega =0$. Could you give a counterexample with non-compact $ M$.

I don't feel like proving Stokes' theorem here. Counterexamples are abundant - just let $ M$ be an open subset of $ \mathbb R^{2}$, and $ w=ydx$.

It is going to be a stupid question,anyway : Stokes says that $ \int_{M}d\omega = \int_{d M}\omega$, why is $ \int_{ d M}\omega = 0$ ?

Oh,yes. I'm not so well used to the notion of a manifold, and it is sometimes misleading, whether a manifold is supposed to be with boundary or without.
Thanks.


calculus
integration
LaTeX
search
articles
complex analysis

Anyway I can place a small  number at the lower right corner of the circle around the integral sign?
\[{ \mathop\oint\limits_{|z|>H}}\frac{dz}{\sqrt[n]{P_n(z)}}\]

I know, it's not standard but there is no notation for identifying precisely which analytic sheet of this multifunction the integration is being performed over.  I could use substack with a $ k=2$ or another number but I guess it would be nice if the integral symbol would do that defining for me.  Maybe it's a dumb idea.  Anyway it could be done though?

Thanks,
Shaw

[url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=comprehensive]The Comprehensive LaTeX Symbol List[/url] (search for \ddashint) shows you how to make your own integral symbols.

Ok Steve.  I've been working on it.  Certainly looks feasible.  I'm just weak with the code.

Thanks!

The closest I can get is to place a minus sign or other math symbol near where I want it but I can't place a $ k$ or make it tiny and if the limits underneath the integral change then that messes up the formatting.  Here's my code taken from the Latex manual I slightly modified.  On my machine in TeXnicCenter, it prints a minus sign (large) where I would want the k.

Can anyone suggest a way to modify it and print the tiny k?

[code]\def\Xint#1{\mathchoice
{\XXint\displaystyle\textstyle{#1}}%
{\XXint\textstyle\scriptstyle{#1}}%
{\XXint\scriptstyle\scriptscriptstyle{#1}}%
{\XXint\scriptscriptstyle\scriptscriptstyle{#1}}%
\!\oint}
\def\XXint#1#2#3{{\setbox0=\hbox{$#1{#2#3}{\int}$}
\vcenter{\vspace{16pt}\hbox{$#2#3$}}\kern-1\wd0}}
\def\anaoint{\Xint-}[/code]

I'd like then to use \anaoint to print the integral below with the $ k$.

\[ \mathop\oint\limits_{|z|>H}\frac{1}{\sqrt[n]{P_n(z)}}dz\]

Here are 2 alternative ways of doing it:
[code]\documentclass{article}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{wasysym} % needed for \ocircle in \anaointb
\DeclareMathOperator*{\anaointa}{\begingroup \displaystyle \unitlength 1pt
 \int\mkern-7.1mu\begin{picture}(0,3)\put(-2.5,3){\circle{6}}\end{picture}
 \hspace{1pt}_{\scriptscriptstyle k}\mkern2mu\endgroup}

\def\Xint#1{\mathchoice
   {\XXint\displaystyle\textstyle{#1}}%
   {\XXint\textstyle\scriptstyle{#1}}%
   {\XXint\scriptstyle\scriptscriptstyle{#1}}%
   {\XXint\scriptscriptstyle\scriptscriptstyle{#1}}%
   \hspace{-10pt}\int}

\DeclareMathOperator*{\anaointb}{\Xint{\ocircle_{\scriptscriptstyle k}}}

\begin{document}
\noindent \textbackslash anaointa\:
\[\anaointa_{|z|>H}\frac{1}{\sqrt[n]{P_n(z)}}dz\]
\textbackslash anaointb:
\[\anaointb_{|z|>H}\frac{1}{\sqrt[n]{P_n(z)}}dz\]
\end{document}[/code]

Steve, thank you.  That's nice code to study.  I worked on it too and don't wish you to think I was just waiting around for others to figure it out.  By chance I came up with this:

[code]
\def\Xint#1{\mathchoice
{\XXint\displaystyle\textstyle{#1}}%
{\XXint\textstyle\scriptstyle{#1}}%
{\XXint\scriptstyle\scriptscriptstyle{#1}}%
{\XXint\scriptscriptstyle\scriptscriptstyle{#1}}%
\!\oint}
\def\XXint#1#2#3{{\setbox0=\hbox{$#1{#2#3}{\oint}$}
\vcenter{\tiny\vspace{16pt}\hbox{$#2#3$}}\kern-1\wd0}}
\def\anaoint{\Xint}
[/code]

and it's perfect even though I don't understand it.  I appreciate yours too; however may I be a little particular?  Hope you don't take offense.  Your circle is a little too large and the k is too small although once I begin studying it I can probably figure out how to change that.

Here's the final display of mine which won't work here:

[code]
\anaoint{k}\limits_{|z|>H}\frac{1}{\sqrt[n]{P_n(z)}}dz=\frac{2\pi i}{\sqrt[n]{z_n}}e^{-2k\pi i/n}
[/code]

Any chance we could get that to work here?  I'd like to ask the question of why the integral has the value based on $ k$.  I realize I could ask anyway, but the expression  looks slick and clearly identifies what sheet is being integrated and the relationship between the sheet,  $ k$,  and the answer.

Actually looks like the circle is ok.  But I think it  looks better if the $ k$ is a little larger and lower down even with the end of the integral sign.  I'll try and figure out how to change your code.

[quote="shawtoos"]Any chance we could get that to work here?[/quote]No chance at all as \def and similar commands cannot be used on the forum for security reasons. Compile the code on your own computer and then you can upload a screen shot of the image.

Ok.  No, no screen shot.  I don't want to rile them over there in Complex Analysis and even just Latex may look like a show-off even though I think it's nice.  Hey, I've made progress with your code:  I've made the circle smaller and used \tiny.  Can you suggest how I might now lower the $ k$ even with the end of the integral sign?

[code]
\int\mkern-7.1mu\begin{picture}(0,3)\put(-2.5,3){\circle{5}}\end{picture}
% \hspace{1pt}_{\scriptscriptstyle k}\mkern2mu\endgroup}
 \hspace{1pt}_{\tiny{k}}\mkern2mu\endgroup}

\def\Xint#1{\mathchoice
   {\XXint\displaystyle\textstyle{#1}}%
   {\XXint\textstyle\scriptstyle{#1}}%
   {\XXint\scriptstyle\scriptscriptstyle{#1}}%
   {\XXint\scriptscriptstyle\scriptscriptstyle{#1}}%
   \hspace{-10pt}\int}
[/code]

Then again, I could just cheat:

\[ \mathop\oint_{\hspace{\minus{}13pt}\substack{\phantom{\big).}k \\ |z|>H}}\frac{1}{\sqrt[n]{P_n(z)}}dz\equal{}\frac{2\pi i}{\sqrt[n]{z_n}}e^{\minus{}2k\pi i/n}\]

still though would like the $ k$ even with the lower end of the integral sign which I can't seem to do with this method.

[quote="shawtoos"][code]
\int\mkern-7.1mu\begin{picture}(0,3)\put(-2.5,3){\circle{5}}\end{picture}
% \hspace{1pt}_{\scriptscriptstyle k}\mkern2mu\endgroup}
 \hspace{1pt}_{\tiny{k}}\mkern2mu\endgroup}[/code][/quote]You haven't been paying attention to the warnings when compiling, which would have told you that you can't use \tiny in math mode. You are restricted to \displaystyle, \textstyle, \scriptstyle or \scriptscriptstyle.

Try this code
[code]\DeclareMathOperator*{\anaointa}{\begingroup \displaystyle \unitlength 1pt
 \int\mkern-7.1mu\begin{picture}(0,3)\put(-2.5,3){\circle{5}}\end{picture}
 \hspace{-3pt}\raisebox{-10pt}{\tiny $k$}\mkern2mu\endgroup}[/code]
\raisebox converts to text mode so you can use \tiny but need to go back to math mode for k. Clearly you can change the numbers in \hspace{-3pt}\raisebox{-10pt} to move the k around.

[quote="shawtoos"]Then again, I could just cheat:
\[ \mathop\oint_{\hspace{ \minus{} 13pt}\substack{\phantom{\big).}k \\
|z| > H}}\frac {1}{\sqrt [n]{P_n(z)}}dz \equal{} \frac {2\pi i}{\sqrt [n]{z_n}}e^{ \minus{} 2k\pi i/n}
\]
still though would like the $ k$ even with the lower end of the integral sign which I can't seem to do with this method.[/quote]
How about
\[ \mathop\oint\limits_{|z|>H}\hspace{\minus{}13pt}\raisebox{\minus{}10pt}{\tiny \(k\)}\hspace{3pt}\frac{1}{\sqrt[n]{P_n(z)}}dz\equal{}\frac{2\pi i}{\sqrt[n]{z_n}}e^{\minus{}2k\pi i/n}\]
which can also be adjusted to whatever you feel is best? Note the use of \(...\) which mean the same as dollar signs but are required on the forum as dollar signs inside a formula are misinterpreted.

[quote="stevem"]You haven't been paying attention to the warnings when compiling, which would have told you that you can't use \tiny in math mode. You are restricted to \displaystyle, \textstyle, \scriptstyle or \scriptscriptstyle.
[/quote]

. . . second time you've caught me doing that . . . guess sometimes I run willy-nilly and hope everything is poke-a-poke  :D .  I know, that's poor (and lazy) programming style; not always I do that though.  The code below locates the "k" perfectly but . . . \tiny is too small and \small is too large:

[code]
\DeclareMathOperator*{\anaointa}{\begingroup \displaystyle \unitlength 1pt
 \int\mkern-7.1mu\begin{picture}(0,3)\put(-2.5,3){\circle{5}}\end{picture}
 \hspace{-3pt}\raisebox{-8pt}{\tiny $k$}\mkern2mu\endgroup}
[/code]

. . . I'll work on it . . .  nice programming exercise :)

[edit]  Ok, \scriptsize is perfect.  Thanks a bunch Steve for helping me.  I think that does it :)

[quote="stevem"]
How about
\[ \mathop\oint\limits_{|z| > H}\hspace{ \minus{} 13pt}\raisebox{ \minus{} 10pt}{\tiny \(k\)}\hspace{3pt}\frac {1}{\sqrt [n]{P_n(z)}}dz \equal{} \frac {2\pi i}{\sqrt [n]{z_n}}e^{ \minus{} 2k\pi i/n}
\]
which can also be adjusted to whatever you feel is best? Note the use of \(...\) which mean the same as dollar signs but are required on the forum as dollar signs inside a formula are misinterpreted.[/quote]

Just changed the raisebox, changed size to scriptsize,  and now I think this is perfect:

\[ \mathop\oint\limits_{|z|>H}\hspace{\minus{}13pt}\raisebox{\minus{}8pt}{\scriptsize \(k\)}\hspace{3pt}\frac{1}{\sqrt[n]{P_n(z)}}dz\equal{}\frac{2\pi i}{\sqrt[n]{z_n}}e^{\minus{}2k\pi i/n}\]

Thanks Steve :)