Let be $ x,y,z\in \mathbb{R}$ such that $ x\plus{}y\plus{}z\equal{}2\ ,\ x^2\plus{}y^2\plus{}z^2\equal{}30\ ,\ x^3\plus{}y^3\plus{}z^3\equal{}116$ . Calculate $ xyz$ .

[quote="tonypr"]Note that $ x^3 \plus{} y^3 \plus{} z^3 \minus{} 3xyz \equal{} (x \plus{} y \plus{} z)(x^2 \plus{} y^2 \plus{} z^2 \minus{} 2(xy \plus{} xz \plus{} yz))$[/quote]

There should be a correction

$ x^3 \plus{} y^3 \plus{} z^3 \minus{} 3xyz \equal{} (x \plus{} y \plus{} z)(x^2 \plus{} y^2 \plus{} z^2 \minus{} (xy \plus{} xz \plus{} yz))$

Proceeding as above gives $ xyz \equal{} 10$ if i haven't made any silly mistake  :maybe:

Ya, I think xyz does equal 10 even though I cannot see any reason tonypr is incorrect.

I think $ x \equal{} 5$, $ y \equal{} \minus{}2$, and $ z \equal{} \minus{}1$, but the order doesn't matter due to symmetry. Therefore, $ xyz \equal{} 10$.


linear algebra
matrix
email
linear algebra open

Given $ A\in C^{n\times n}$ and a positive semidefinite
  Hermitian $ Q\in C ^{n\times n}$. How to find a Hermitian
  solution $ X$ such that $ (X \minus{} A)(X \minus{} A)^* \equal{} Q$?
If you have  a solution, please email me: miwalin@163.com

Write $ Q\equal{}UDU^*$ with $ U$ unitary matrix and $ D$ a diagonal matrix with nonnegative real entries, then $ X\equal{}A\plus{}U\sqrt{D}$ does it.

Hi hjbrasch,
your $ X$ is not hermitian !


geometry
3D geometry
number theory
relatively prime

[quote="surge"]If $a$ and $b$ are natural numbers and $5a^3 = 6b^4$, what is the smallest possible value of $a + b$?

Please show work.[/quote]

[hide]Basically logic I think.

So $b$ has to be a multiple of 5 and $a$ a multiple of 6.

Through some obscure thinking of mine I got $a+b=360$ which seems too big to be right. :( If it is though, I might post my obscure thinking. :!: [/hide]

[hide]a has to be a multiple of 6 and b has to be a multiple of 5. Since they are relatively prime we can solve the problem 1 factor at a time as they will not intermesh and interfere with each other.

Assume a = 6 and b = 5. Then we have

5 * 6^3 = 6 * 5^4

You can only add a factor to the left side as a cube, and you can only add a factor to the right side as a fourth power. Since the lhs needs a 5^3 to balance out the factos of 5, we do so and let a = 5 * 6 = 30 and get

5 * 5^3 * 6^3 = 6 * 5^4

Notice that the lhs will have a factor of 6 raised to some power that is a multiple of 3, and that the rhs will have a factor of 6 raised to some power that is 1 more than a multiple of 4. The smallest would be 9, and we can get that by letting b = 5 * 6^2 = 180 and a = 5 * 6^3 = 1080

We would then have

5 * (5 * 6^3)^3 = 6 * (5 * 6^2)^4

5 * 5^3 * 6^9 = 6 * 5^4 * 6^8

5^4 * 6^9 = 5^4 * 6^9

Hence,

[b][u]a = 1080
b = 180
a + b =1260[/u][/b][/hide]

I remember I did this problem in a matter seconds but I forgot what I did though since it was on my calculator but it certainly wasn't as complicated as lasagnaman's solution. I'll post it when I remember what I did. :surf:

[quote="surge"]If $a$ and $b$ are natural numbers and $5a^3 = 6b^4$, what is the smallest possible value of $a + b$?

Please show work.[/quote]

[hide]We can show that a is of the form $5^{x}*6^{y}$, and b is of the form $5^{z}*6^{w}$.

So by laws of exponents

$5^{3x+1}*6^{3y}=5^{4z}*6^{4w+1}$

So $3x+1=4z$ and $3y=4w+1$

We have to find the smallest values of x, y, z, and w that work.

In this case it is x=1, z=1, y=3, w=2.

So $a=5*6^3$ and $b=5*6^2$

So a=1080 and b=180

So a+b=1260

$1260$[/hide]

Oh duh I know what I did wrong.

[hide]So we know that $a$ is a multiple of 6 and $b$ is a multiple of 5.

So we can write that $a=6x$ and $b=5y$.

$5a^3=6b^4\implies\frac{a^3}{b^4}=\frac65\implies \frac{216x^3}{625y^4}=\frac65$

So $\frac{x^3}{y^4}=\frac{125}{36}$

Also, note that $\frac{125}{36}=\frac{5^3}{2^23^2}$, so we must find a number that when multiplied by both the numerator and denominator leaves the top as a perfect cube and the bottom as a perfect 4th power.

Guess and check yields $2^63^6$ so $x=180, y=36$.

So then, plugging back into $a,b$, we get $a+b=1080+180=\boxed{1260}$ [/hide]


Show that an integer can be expressed as the difference of two squares if and only if it is not of the form $4k+2 \; (k \in \mathbb{Z})$.

$ (x^2\minus{}y^2)\equal{}m$
$ (x\minus{}y)(x\plus{}y)\equal{}m$
Because $ x\minus{}y\equiv x\plus{}y(\mod 2)$ so $ m$ is not in the form $ 4k\plus{}2$
Now we prove that if m is not in the form the equation has solution.
[u]Case 1[/u] $ m\equiv 1(\mod 2)$
Suppose $ m\equal{}pq$ then
$ x\equal{}\frac{p\plus{}q}{2},y\equal{}\frac{p\minus{}q}{2}$ is solution.
[u]Case 2[/u]$ m\equiv 0(\mod 4)$
$ m\equal{}4pq$ then
$ x\equal{}{p\plus{}q},y\equal{}p\minus{}q$ is solution.
Other problem:
Find the number of pair $ (x,y)\in N$ $ x^2\minus{}y^2\equal{}m$

If $a^2-b^2=m$, then $(a+b)(a-b)=m$, where the two factors have the same parity, hence either $m$ is odd or $4|m$.
And note that $(n+1)^2-n^2=2n+1$ gives all the odd numbers and $(n+1)^2-(n-1)^2=4n$ generates all integers divisible by $4$.


[quote="GameBot"]What is the 100th digit after the decimal point of the decimal representation of 1/7?[/quote]

same as 4th digit, which is 8.

$ \frac{1}{7} \equal{} 0.\overline{142857}$

So it repeats every $ 6$ digits.

Since we are looking for the $ 100^{\text{th}}$, we must find the remainder of $ 100 \div 6$.  This is $ 4$.

So the $ 100^{\text{th}}$ digit is the same as the fourth, so the answer is $ \boxed{8}$.

[quote="BOGTRO"]Please don't spam. If a question has already been answered, don't answer it again unless there is something new you are posting.[/quote]

1.  You're spamming by telling me not to spam.
2.  My post wasn't spam.  In fact, yours was.  Your post didn't clearly explain your method.
3.  If you're trying to get me back from HSB, it isn't working.


How many even three-digit integers have the property that their digits, read left to right, are in strictly increasing order?

$ \textbf{(A) } 21 \qquad \textbf{(B) } 34 \qquad \textbf{(C) } 51 \qquad \textbf{(D) } 72 \qquad \textbf{(E) } 150$

I just listed them all out really quickly and got [hide]B[/hide]

How about considering each case of an even digit.  Clearly 0 and 2 are impossible if they are to be in strictly increasing order.  So if the last digit is 4 then there are 3C2 ways to pick to strictly less than it.  So similary with 6 and 8, we can write 3C2 + 5C2 + 7C2= 34

Yeah, I believe that's what I did, except the $\binom{n}{2}$ sums are the sums of the first two, four, and six positive integers, so you just need to add those.

Yea...i Just listed them too and got [hide]B[/hide]

[hide]You can do casework based on the first number, if it's in the form 1AB, then:

-if A=2, $B\in \{4,6,8\}$ (3 possibilities)
-if A=3, 3
-A=4, 2
-A=5, 2
-A=6, 1
-A=7, 1

For a total of 3+3+2+2+1+1=12. With 2AB, it's just 12-3, then 12-3-3, 12-3-3-2, 12-3-3-2-2, 12-3-3-2-2-1, with sum 34.[/hide]

Yep, casework did the trick for me. I skipped right over the 'even' part at first and would have gotten this wrong had 84 been one of the choices.

This was just a case based problem
List out all the cases of digits ending with 0 2 4 6 8
As 0 and 2 cannot be there so
No ending with 4
Just select any of the four numbers below it and arrange them in one order which 4c2 but some cases which include 0 in front is 01,02,03
Hence total ways is 4c2-3
You can see a pattern
For 6 and 8 which is
6c2-5 ,8c2-7
Adding them up gives 34 as desired


Find all primes p and q such that
p 2 -p+1=q^3

Neglected but intriguing problem  :)

A shortened solution using a lot of algebra (so not too nice  :? ) :

rewriting p(p-1) = (q-1)(q+q+1) gives q = k(p-1)+1 for some k in N.

replacing, factoring, ... gives :

p(k:^3:(p-2)+3k-1) = -k(3-3k+k)

LHS is > 0 so k = 1,2 or 3.

Now it is easy, if I'm not wrong only (19,7) is solution.

Does any one has a nicer solution ?


$ a^2\minus{}b^2 / a$
$ a^2\minus{}b^2 / b$
Then prove:
$ |a^2\minus{}b^2|\equal{}1$

This is - kind of - a question of definition; if we allow the statement $ 0|0$ (which makes sense since 0 is a multiple of 0) could just as well have the case $ a\equal{}b\equal{}0$.

[hide="Otherwise"]$ \frac{1}{a\plus{}b} \equal{} \frac{a\minus{}b}{a^2\minus{}b^2} \equal{} \frac{a}{a^2\minus{}b^2} \minus{} \frac{b}{a^2\minus{}b^2} \in \mathbb{Z}\\\implies a\plus{}b \equal{} \pm 1 \implies \operatorname{gcd}(|a|,|b|) \equal{} \operatorname{gcd}(|a|,a\plus{}b) \equal{} 1$

However $ (a^2\minus{}b^2)|a$ and $ (a^2\minus{}b^2)|b\implies (a^2\minus{}b^2)|\operatorname{gcd}(a,b) \equal{} 1$.[/hide]


Find the maximum real number M satisfying $ \frac{a^2\plus{}(bc)^2}{(a\plus{}b\plus{}c)^2}$≥$ M$ where a,b,c are 3 sides of a triangle ABC(max{A,B,C}>90 degrees)

[quote="SUPERMAN2"]Find the maximum real number M satisfying $ \frac {a^2 \plus{} (bc)^2}{(a \plus{} b \plus{} c)^2}$≥$ M$ where a,b,c are 3 sides of a triangle ABC(max{A,B,C}>90 degrees)[/quote]
$ M \equal{} 0$... Consider the triangle where $ a \equal{} 0$ and $ b \equal{} c \equal{} x$ Then we have $ \frac {x^4}{4x^2} \equal{} \frac {x^2}{4} \ge M$. Then just make $ x$ abritary small.

[quote="Umut Varolgunes"]Your example does not fit in with the condition that triangle $ ABC$ is obtuse angled.[/quote]
Just let $ c$ be slightly bigger than $ b$ and let $ a$ be slightly bigger than $ 0$ ;)

There are two typing mistakes in the statement of this problem.

It should read: Find the largest real M such that   $ a^2 +b^2 +c^2 >M(a+ b  +c) ^2$   where   a,  b  ,and c are the sides of an obtuse triangle

The inequality is strict: >      Not $ \geq$

 Hint [hide]Start by proving that $ \frac{1}{3}$ $ \leq$ M $ \leq$ 1.[/hide]


why cant i make f(x)=x^{4} - 8x^{3} + 18x^{2} - 12x + 24
into latex it gives me
this error
$ f(x) \equal{} x^{4} \minus{} 8x^{3} \plus{} 18x^{2} \minus{} 12x \plus{} 24$

sorry if this is wrong forum but i didnt know which one to put since this was questions too.  anyways i just put $ \$$ around it and it gives me that latex dangerous error

Now it should work. It was a small glitch on our end. Thanks for letting us know.


Find all $f:N_{0} -> N_{0}$ which satisfy the equation
\[ f(f(m)^{2} + f(n)^{2}) = m^{2} + n^{2} \]
for all $m,n \in N_{0}$.

There is only one function, namely $f(n)=n$. Firstly, observe that $f$ is injective. Secondly, let's find $f(0)$ and $f(1)$. Put $m=n=0$, then $f(2f^2(0))=0$. Set $n=m=2f^2(0)$, so $f(0)=8f^4(0)$ or $f(0)=0$. Let $n=1$ and $m=0$, hence $f(f^2(1))=1$. Put $n=f^2(1)$ and $m=0$, therefore, $f(1)=f^4(1)$ or $f(1)=1$, since $f$ is one-to-one mapping. Now we can determine that $f(2)=2$, $f(4)=4$, $f(5)=5$, $f(8)=8$ with putting $n=m=1$, $n=2$ and $m=0$, $n=2$ and $m=1$, $n=m=2$ respectively. It is well known that $5^2+0^2=3^2+4^2$. That implies that $f(f^2(0)+f^2(5))=f(f^2(3)+f^2(4))$ or $f^2(0)+f^2(5)=f^2(3)+f^2(4)$ since $f$ is injective. Therefore $f(3)=3$. With the same technique we get $f(9)=9$, $f(10)=10$, and, furthermore, $f(6)=6$, $f(7)=7$ because $6^2+8^2=10^2$ and $7^2+1^2=5^2+5^2$. For other numbers, reason over the following equalities: $(2t+2)^2+(t-4)^2=(t+4)^2+(2t-2)^2$ for odds and $(2t+1)^2+(t-2)^2=(2t-1)^2+(t+2)^2$ for evens.


algebra
polynomial
function
vector
arithmetic sequence
absolute value
advanced fields

I understand that it would be very important if the prime number
sequence was found or even a subsequence of $p_n$.
What work has been done in this direction ?
What is the trend and what are the tools to work about this problem?

The only relevant result that comes to my mind is the Matiyasevich theorem that states that there exists a polynomial of many (23?) variables with integer coefficients, whose set of positive values on integer arguments coincides with the set of primes (of course, for most choices of arguments, the value of this polynomial is negative).

What people try to do is rather to ask some questions about various properties of the sequence $p_n$ (like twin prime conjecture, still unsettled, or arbitrarily long arithmetic progression conjecture, settled recently by Green and Tao, etc.).

could i please have some more information on this polynomial ?

ZetaX where does this polynomial come from(what is behind it) ?
why is it so ugly?
it seems there is no type of order in it.
fedja told me that the values it can take are all the primes,
so why is there still need for riemann zeta function to find out
distribution of primes when we could solve that problem with this
polynomial ?

an observation

from what i knew from a friend and own research on the net
i understood that most formulas are computational infeasable
for producing primes.
also what i notice for this polynomial is that it can be thought
as a function on $R^{n}$ where n is the number of variables.
now we try to find distinct primes with the polynomial but
it seems that we can get collisions...like the injectivity of the
function could be compromised by the fact that for more vectors
in $R^{n}$ we could get the same prime.
i don't know if i'm right,but is this the computational infeasable
reason in this case of generating primes tghrough this polynomial ?

The polynomial comes from the set of equations given here: http://mathworld.wolfram.com/PrimeDiophantineEquations.html
And the problem is to find $a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t,u,v,w,x,y,z$ such that it gets positive (try ;) ).

It's a german page, but click the list of links at the bottom: http://www.beweise.mathematic.de/primenumbers-jsww.html
(the existance shows constrictively that the primes are a diophantine set)

http://mathworld.wolfram.com/Prime-GeneratingPolynomial.html

ZetaX there they say that this polynomial
"The best-known polynomial that generates (possibly in absolute value) only primes is
$P(n)=n^2+n+41$".
but if $41|P(41)$ how can this be a prime genereating [b]only[/b] polynomial ?

here http://www.recmath.org/contest/description.php
i read that $P$ has only 40 prime values.
(how about the other questions above?)


USAMTS email LaTeX conics ellipse vector

How do I download fancyhdr.sty?  When I click on the link from the usamts website to download it, it gives me a program of some sort, but no way to download it.  The first few lines are:

% fancyhdr.sty version 3.0
% Fancy headers and footers for LaTeX.
% Piet van Oostrum,
% Dept of Computer and Information Sciences, University of Utrecht,
% Padualaan 14, P.O. Box 80.089, 3508 TB Utrecht, The Netherlands
% Telephone: +31 30 2532180. Email: piet@cs.uu.nl
 :huh:

If you are using Miktex, go to Start --> Programs --> Miktex --> browse packages.  There will be a huge list of packages and all you need to do is look for fancyhdr.  When you find it, right click it and hit install.  If you can't install it, try turn off your firewall and any internet security (they prevented me from installing it)

Phoenix257: So installing fancyhdr worked in the end for you then?

I do have fancyhdr installed now, but why do the following programs have nearly twenty error messages each?

I don't see why the templates from the USAMTS website should have that many errors.

Note:  they are the USAMTS answer and include files, but there is nothing in them that is directly related to the solutions.

[quote="roadnottaken"]I do have fancyhdr installed now, but why do the following programs have nearly twenty error messages each?[/quote]Each? You only compile one file - the solutions file. But you must take care not to change the name of the include file.

You may call the USAMTS_solutions_forma.tex whatever you like. BUT change the name of USAMTS_include.tex to usamts.tex otherwise LaTeX won't be looking for the right file to include.

You only need to open and compile USAMTS_solutions_forma.tex. Be aware that until you add your own text to this file no pdf output will be created.

To summarise:
1. Make sure that USAMTS_solutions_forma.tex and usamts.tex are in the same directory.
2. Open USAMTS_solutions_forma.tex and put your own stuff where it tells you.
3. Compile USAMTS_solutions_forma.tex to get USAMTS_solutions_forma.pdf
4. Any errors are listed in USAMTS_solutions_forma.log.

If you are still getting errors then you need to tell us what they are.

I have one more question:  what is the simplest way to make a picture of specific shapes for use in a $\LaTeX$ document.  I have in mind the shapes that correspond to an exact equation.

If you just mean graphics then the LaTeX picture command is very powerful. See http://www.artofproblemsolving.com/Forum/viewtopic.php?p=706387 and Chapter 5 of http://tobi.oetiker.ch/lshort/lshort.pdf

To set up a bezier curve, the command \qbezier$(x_{1},y_{1})(x_{2},y_{2})(x_{3},y_{3})$ is used.  Could someone please explain [b][i]exactly[/i][/b] how the $(x_{2},y_{2})$ is determined to make a circle and an ellipse, using four bezier curves each?  The circle and ellipse need only be approximate.

I think the piece of information that you need is that in [i]\qbezier(a,b)(c,d),(e,f)[/i], (c,d) is defined to be the intersection of the tangents to the curve at (a,b) and (e,f).

I'm not saying it's easy now but you could try estimating the middle point from the other two then adjusting by trial and error.

Well, somebody wrote a Java program to do draw circles with 4 Bezier curves. But I couldn't get the program to compile properly using the GNU Java compiler, so I just use MetaPost to draw circles. You can just use the [url=http://www.tlhiv.org/cgi-bin/MetaPostPreviewer/index.cgi]MetaPost previewer[/url] and export it to .eps when you are done. Then it is really easy to include into LaTeX documents. By the way, a metapost circle is done like this:

[code]
draw fullcircle;
[/code]

If you want to change the diameter, you can write

[code]
draw fullcircle scaled Xcm;
[/code]

where X is the length of the diameter. You can also use other units like mm, cm, in, etc. Also, if you don't want it to be centered in the middle of the screen you will have to shift it like so:

[code]
draw fullcircle scaled Xcm shifted (x1,y1);
[/code]

where (x1,y1) is the vector to shift it by.

Also, you will certainly find [url=http://tex.loria.fr/prod-graph/zoonekynd/metapost/metapost.html]these MP examples[/url] to be helpful.

For ellipses you will want to look at [url=http://www.tlhiv.org/MetaPost/tutorial/tutorial.html#CircleParametrization]this[/url].

Yes, I am using this topic for any and all questions I have about $\LaTeX$. :wink:

Does anyone know how to make a page-break in $\LaTeX$?


inequalities
calculus
inequalities proposed

$ a,b\in\mathbb{R^\plus{}}\cup\{0\}$. Prove that

[b]a)[/b] If $ 0<p\le1$ then
          $ (a\plus{}b)^p\le a^p\plus{}b^p$

[b]b)[/b] If $ p>1$ then
          $ (a\plus{}b)^p\le2^{p\minus{}1}(a^p\plus{}b^p)$

Actually both a) and b) and be solved using calculus .
Let a/(a+b)=m , then b/(a+b)=1-m
     a) becomes 1≤m^p+(1-m)^p
     R.H.S. of a) becomes m^p+(1-m)^p
     Differentiate R.H.S. with respect to m and we get the minimum value of R.H.S. attains when m=1-m , i.e m=1/2 .
   Substitute m=1/2 into them and both a) and b) are proved.


calculus
integration
trigonometry
calculus computations

I need some help with this: \[ \int_{0}^{\infty} x^3 e^{\minus{}ax^2}\sin{bx} dx\]
where a>0 , b>0

$ I\equal{}\left\{\sin\,bx\int\;x^3e^{\minus{}ax^2}\;dx\right\}\;\bigg|_0^\infty\minus{}\int_0^\infty\;\left\{b\,\cos\,bx \int x^3e^{\minus{}ax^2}\;dx\right\}\;dx$

now $ \int\;x^3e^{\minus{}ax^2}\;dx\equal{}\frac{1}{2a^2}\int\;ye^{\minus{}y}\;dy\qquad\text{where}\qquad y\equal{}ax^2$

and $ \int\;ye^{\minus{}y}\;dy\equal{}\minus{}(y\plus{}1)e^{\minus{}y}\plus{}C\equal{}\minus{}(ax^2\plus{}1)e^{\minus{}ax^2}\plus{}C$

so,
 $ I\equal{}\underbrace{\minus{}\frac{1}{2a^2}\left\{\sin\,bx\;(ax^2\plus{}1)e^{\minus{}ax^2}\right\}\;\bigg|_0^\infty}_{\equal{}\quad 0}\;\;\plus{}\;\;\frac{b}{2a^2}\int_0^\infty\;\left\{\cos\,bx\; (ax^2\plus{}1)e^{\minus{}ax^2}\right\}\;dx$

$ \equal{}\frac{b}{2a}\int_0^\infty\;\cos\,bx\;x^2e^{\minus{}ax^2}\;dx\plus{}\frac{b}{2a^2}\int_0^\infty\;\cos\,bx\; e^{\minus{}ax^2}\;dx\qquad \ldots (1)$

do you know $ DUIS$ ? using that you can show

$ \int_0^\infty\;\cos\,bx\; e^{\minus{}ax^2}\;dx\equal{} \sqrt{\frac{\pi}{4\,a}}\;e^{\minus{}\frac{b^2}{4\, a}}$

and
$ \int_0^\infty\;\cos\,bx\;x^2e^{\minus{}ax^2}\;dx \equal{} \minus{}\;\frac{d^2}{db^2} \int_0^\infty\;\cos\,bx\; e^{\minus{}ax^2}\;dx$

plug these $ 2$  results in $ (1)$

Or, we can just integrate with respect to $ a$ and $ b$, use $ \int_0^\infty\;\cos\,bx\; e^{ \minus{} ax^2}\;dx \equal{} \sqrt {\frac {\pi}{4\,a}}\;e^{ \minus{} \frac {b^2}{4\, a}}$ and differentiate again with respect to $ a$ and $ b$. In other words:

$ \int_{0}^{\infty} x^3 e^{ \minus{} ax^2}\sin{bx} dx \equal{} \sqrt {\frac {\pi}{4}}\frac { \partial^2}{\partial a\,\partial b}\left(\frac {1}{\sqrt {a}}e^{ \minus{} \frac {b^2}{4a}}\right)\equal{} \sqrt{\frac{\pi}{a}}\frac{b\, (6a\minus{}b^2)}{8 a^3} e^{\minus{}\frac{b^2}{4a}}$

Looks good, big thanks. Can u also show that \[ \int_0^\infty\;\cos\,bx\; e^{ \minus{} ax^2}\;dx \equal{} \sqrt {\frac {\pi}{4\,a}}\;e^{ \minus{} \frac {b^2}{4\, a}}\]


linear algebra
matrix
algebra
function
domain
geometry
geometric transformation

Suppose $ F: \mathbb{R}^3 \to \mathbb{R}^3$ is a diffeomorphism, and that the Jacobian matrix of $ F$ is orthogonal at each point of $ \mathbb{R}^3$.  Is $ F$ an isometry?  How would you prove that?

One of Liouville's theorems states that if $ F$ is a conformal map on a connected open subset of $ \mathbb R^n$ with $ n>2$ (i.e. $ DF(x)^T DF(x)$ is a scalar matrix for all $ x$), then $ F$ is a Mobius transformation. With the domain being $ \mathbb R^n$, you get only rotations, reflections, and rescaling.

But you don't need Liouville's theorem here. Note that $ |F(a)\minus{}F(b)|\le |a\minus{}b|$ (by integrating along the segment $ [a,b]$), and the same holds for $ F^{\minus{}1}$.


we have to prove that $\frac{7}{27}. (x+y+z)^3 + 2xyz \ge (xy+yz+zx)(x+y+z)$

but ths is the same that prove $7\sum{a^3} + 5\sum{xyz} \ge 12\sum{a^2b}$.

by schur, we have $5\sum{a^3} + 5\sum{xyz} \ge 10\sum{a^2b}$

and by bunching, $2\sum{a^3} \ge 2\sum{a^2b}$.

so, the problem is solved.  :)

What if we have an additional condition, which says that x, y, z are lengths of three sides of a triangle.
Will we have an stronger up limit

[quote="e.lopes"]

by schur,
[/quote]What's schur?

I guess it's cauchy schwarts ineq    :ninja:

Obviously not.  Anyways, schur states that

$x^t(x-y)(x-z)+y^t(y-x)(y-z)+z^t(z-x)(z-y)\geq 0$ for all $x,y,z\geq 0$ and all $t\in\mathbb{R}$.

The case in point here was for $t=1$.

The sign in your last inequality is flipped.

[quote="mathematica"]Prove that $xy+yz+xz-2xyz \leq \frac{7}{27}$ when $x+y+z=1$.[/quote]

First don't forget to require x, y, z to be $\geq 0$ (else it is wrong), then please continue discussion at http://www.mathlinks.ro/Forum/viewtopic.php?t=50902 .

  Darij


Euler
geometry
trigonometry
circumcircle
trapezoid
geometry unsolved

Show that Euler line of a nonequilateral $ \triangle ABC$ is tangent to its incircle $(I),$ if and only if the following relation holds:

\[ \sqrt {1 \minus{} 8\cos A \cos B \cos C} \equal{} \frac {| (\sin A \minus{} \sin B)(\sin B \minus{} \sin C)(\sin C \minus{} \sin A) |}{(1 \minus{} \cos A)(1 \minus{} \cos B )(1 \minus{} \cos C)} \]

Standard notation: $ (O, R)$ is the circumcircle, $ (I, r)$ the incircle, $ H$ the orthocenter, $ a = BC, b = CA, c = AB$ and $ p$ is the triangle semiperimeter. In addition, $ K, L, M$ are midpoints of $ BC, CA, AB;$ $ D, E, F$ are incircle tangency points with $ BC, CA, AB;$ $ U, V, W$ are feet of the altitudes $ AH, BH, CH;$ the internal angle bisectors $ AI, BI, CI$ cut the circumcircle again at $ X, Y, Z.$

Power of $ H$ to $ (O)$ is $ HO^2 - R^2 = 2\ \overline{HU} \cdot \overline{HA} = - 8R^2 \cos A \cos B \cos C$ $ \Longrightarrow$ the LHS is equal to $ \frac {HO}{R}$. Using the result of [url]http://www.mathlinks.ro/viewtopic.php?t=256552[/url] (also proved at [url]http://www.mathlinks.ro/viewtopic.php?t=260863[/url]), the RHS is equal to

$ \frac {|(b - c)(c - a)(a - b)|}{8R^3 \cdot \prod (1 - \cos A)} = \frac {|\overline{KD} \cdot \overline{LE} \cdot \overline{MF}|}{KX \cdot LY \cdot MZ} = \frac {|\overline{DU} \cdot \overline{EV} \cdot \overline{FW}|}{r^3} =$

$ = \frac {|(b - c) (p - a) \cdot (c - a)(p - b) \cdot (a - b)(p - c)|}{abc \cdot r^3} =$

$ = \frac {|(b - c)(c - a)(a - b)| \cdot |\triangle ABC|^2}{4R \cdot |\triangle ABC| \cdot r^3 \cdot p} = \frac {|(b - c)(c - a)(a - b)|}{4Rr^2}$

$ \Longrightarrow$ the given trigonometric condition is equivalent to $ HO = \frac {|(b - c)(c - a)(a - b)|}{4r^2}$. The Euler line $ OH$ is tangent to the incircle $ \Longleftrightarrow$ $ |\triangle IOH| = \frac {_r}{^2}\cdot HO.$ It is therefore sufficient to show that for arbitrary $ \triangle ABC,$ $ |\triangle IOH| = \frac {|(b - c)(c - a)(a - b)|}{8r}$. WLOG, assume $ a > b > c.$ This area can be calculated from areas of the trapezoids $ \square KOHU, \square KOID, \square DIHU.$

$ |\triangle IOH| = |\square KOHU| - |\square KOID| - |\square DIHU| =$
$ = \frac {_1}{^2}(OK + HU) \cdot KU - \frac {_1}{^2}(OK + ID) \cdot KD - \frac {_1}{^2}(ID + HU) \cdot DU =$
$ = \frac {_1}{^2} (OK \cdot DU + HU \cdot KD - ID \cdot KU) =$

$ = \frac {_1}{^2} [R \cos A \cdot \frac {(b - c)(p - a)}{a} + 2R \cos B \cos C \cdot \frac {b - c}{2} - R(\cos A + \cos B + \cos C - 1) \cdot \frac {b^2 - c^2}{2a}] =$

$ = R\ \frac {b - c}{4a} ((b + c - a) \cos A + 2a \cos B \cos C - (b + c)(\cos A + \cos B + \cos C - 1)) =$

$ = R\ \frac {b - c}{4a} (a(2 \cos B \cos C - \cos A) - (b + c)(\cos B + \cos C - 1)).$

Substituting:

(1) $ b \cos C + c \cos B = a,$

(2) $ R (a \cos A + b \cos B + c \cos C) = 2|\triangle ABC|,$

(3) $ 2aR (\cos A + \cos B \cos C) = ah_a = 2|\triangle ABC|\ \Longrightarrow$

$ |\triangle IOH| = R\ \frac {b - c}{4a} (b + c + a - 2a(1 + \cos A)) =$ $ R\ \frac {b - c}{2a} (\frac {|\triangle ABC|}{r} - \frac {a \sin A}{\tan \frac {A}{2}}) =$

${ = \frac {b - c}{8r} (bc - 2a(p - a)}) = \frac {|(b - c)(c - a)(a - b)|}{8r}$

[color=red]EDIT:[/color] I should have searched earlier. See [url]http://www.mathlinks.ro/viewtopic.php?t=2878[/url].

Thanks for your very nice solution yetti!. I also used the result of the topic [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=2878]An IOH smart and interesting problem[/url]. Incircle $(I,r)$ is tangent to the Euler line $\Longleftrightarrow$ $ | \triangle IOH| \equal{} \frac {_1}{^2}r \cdot OH.$ Hence

$ \frac {r^2 \cdot OH^2}{4} \equal{} | \triangle IOH|^2 \equal{} \frac {(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2}{64r^2}$

Now, we use $ OH^2 \equal{} 9R^2 \minus{} (a^2 \plus{} b^2 \plus{} c^2),$ together with the identities

$ a \equal{} 2R \cdot \sin A \ , \ b \equal{} 2R \cdot \sin B \ , \ c \equal{} 2R \cdot \sin C$

$ \frac {9}{4} \minus{} \sum \sin^2 A \equal{} \frac {R^4}{r^4}(\sin A \minus{} \sin B)^2 \cdot (\sin B \minus{} \sin C)^2 \cdot (\sin C \minus{} \sin A)^2 \equal{}$

$ \frac {1}{4} \minus{} 2\cos A \cdot \cos B \cdot \cos C \equal{} \frac {R^4}{r^4}(\sin A \minus{} \sin B)^2  (\sin B \minus{} \sin C)^2  (\sin C \minus{} \sin A)^2$

In the $\text{RHS}$ we use the identity $ \frac {r}{R} \equal{} 4\sqrt {\frac {_1}{^8}(1 \minus{} \cos A)(1- \cos B)(1-\cos C)}$

After extracting the square root in  both sides of the equation, we obtain

$ \sqrt {1 \minus{} 8\cos A \cos B \cos C} \equal{} \frac {| (\sin A \minus{} \sin B)(\sin B \minus{} \sin C)(\sin C \minus{} \sin A)|}{(1-\cos A)(1-\cos B) (1 \minus{} \cos C)}$


i wonder how many people live in north america... :huh:

like thousands of other users, i also live in north america.

As of 2001, an estimated 454,225,000 people live in North America.

I live in North America now, but I've lived in Japan for 15 years :)

apparently aops has a overwhelming number of north american people gee who would of thought

[quote="h78reg"]any Australian users on AOPS?[/quote]

I know that TripleM is from New Zealand.

I'm suprised there haven't been any votes from Europe.

i live in antartica, no just joking. i live in na, but it is potentially useless info

I live in North America like almost everyone else.

[quote="h78reg"]there doesn't seem to be a lot of international participation in the poll.[/quote]

I don't think a lot of the international people visit fun and games.

yea, i think they go on the olympiad and pre-olympiad sections more often.

[quote="h_s_potter2002"]I don't think a lot of the international people visit fun and games.[/quote]

Yes, it's probably easier for international students (even those who are fluent in English) to understand math than well...F&G stuff.

Um... I don't know how to answer where I live because every year I go to Brasil and I have a house there, but I have a house here in Texas too. Sometimes I spend equal amounts of time in each place over the year, but one year we stayed in Brasil for about 8 months. I do have a slight southern twang, but it's not very noticable, I don't have an accent or anything. But I sounded british after 6 months in Brasil because of the accent when people speak english sometimes! I have a normal voice that is low for my age, (actually just plain low) but occaisionally when I'm being careless you'll hear me say 2-3 words with the texas draw. I have a joke of mine (there it is!) , "Do I tawk Texin?"  :rotfl:  I did not do that on purpose! It was an accident, but now I make fun of myself with it. My mom does it too sometimes, I even heard my dad (born and raised in Brasil) do it once. Yeah... There was this website that said, "Except for Texans! They're the only ones i have known to put three syllables in "hat"" It's true... The bank lady says, "Turn up back right, then take a little left to the corner and drive on up back to the big ol' trees and turn back right."  :rotfl:  Home sweet home, I don't like country music but...  "coz i'ma red necked woman!"  :rotfl:  Texas for ya! (I should say ya'll, but I will refrain myself from that embarrassment :lol: )

i lived in china for 9 yrs, texas for 1.5, and now maryland for 7. I can't really speak chinese, and according to my friends i have a really slight, indistinguishable accent, which is probably a mixture of cantonese(native lang) and texan drawl (where i learned english). I use "y'all" regularly, but nobody in maryland seems to care.

Asia; more precisely, Taiwan. However I might be moving to North America for college next year...

:? there seem to be a lot higher percentage of users that are not american than is depicted in this poll

I live in Northern Virginia, near Washington D.C.

I live in carmel, close to Indianapolis.  Well A lot of people go to Aops are from North America because the Aops spy center/headquarters is here.


A question:
From $ 1,2,3,.....,50$ we select $ 25$ numbers and another $ 25$ numbers from $ 51,52,53,.....,100$ .But,the difference of any two  in these $ 50$ numbers is not allowed to be $ 0$ or $ 50$.what is the sum of these $ 50$ numbers?

 :roll:  :oops:

[hide]
Consider the $ 50$ numbers $ \mod 50$. Since no two numbers can have a difference of $ 0\mod 50$ all numbers must be distinct.

So we have $ 1\plus{}2\plus{}3\plus{}4...\plus{}50 \equal{} \frac{50(51)}{2} \equal{} 1275$

But to any 25 of these numbers, we must add $ 50$

So the total is $ 1275\plus{} 25(50) \equal{} 2525$[/hide]


calculus
integration
trigonometry
symmetry
function
calculus computations

It is. Note that the cosine and tangent terms integrate to zero by symmetry, so we have

$ 2\int_0^{\frac{\pi}2}(\sin x)^{1/2009}\,dx.$

This can be expressed as $ \frac{\Gamma\left(\frac12\right)\Gamma\left(\frac{2010}{4018}\right)}{\Gamma\left(\frac{4019}{4018}\right)}$

The general formula there is that

$ \int_0^{\frac{\pi}2}\cos^ax\sin^bx\,dx\equal{}\frac{\Gamma\left(\frac{a\plus{}1}2\right)\Gamma\left(\frac{b\plus{}1}2\right)}{2\Gamma\left(\frac{a\plus{}b\plus{}2}2\right)}.$

[quote="kunny"]Evaluate
\[ \int_0^{\pi} (\sqrt [2009]{\cos x} \plus{} \sqrt [2009]{\sin x} \plus{} \sqrt [2009]{\tan x})\ dx.\]
[/quote]

Let $ \ I\equal{}\int_0^{\pi} (\sqrt [2009]{\cos x} \plus{} \sqrt [2009]{\sin x} \plus{} \sqrt [2009]{\tan x})\ dx$

$ \implies I\equal{}\int_0^{\frac{\pi}{2}} (\sqrt [2009]{\cos x} \plus{} \sqrt [2009]{\sin x} \plus{} \sqrt [2009]{\tan x})\plus{} (\sqrt [2009]{\cos (\pi\minus{}x)} \plus{} \sqrt [2009]{\sin (\pi\minus{}x)} \plus{} \sqrt [2009]{\tan (\pi\minus{}x)})\ dx$

$ \implies I\equal{}\int_0^{\frac{\pi}{2}} (\sqrt [2009]{\cos x} \plus{} \sqrt [2009]{\sin x} \plus{} \sqrt [2009]{\tan x})\plus{} (\minus{}\sqrt [2009]{\cos x} \plus{} \sqrt [2009]{\sin x} \minus{} \sqrt [2009]{\tan x})\ dx$

$ \implies I\equal{}2\int_0^{\frac{\pi}{2}} ( \sqrt [2009]{\sin x} )\ dx$

$ \implies I\equal{}2\int_0^{\frac{\pi}{2}} ( sin x )^{\frac {1}{2009}}\ dx$

Now apply gamma function


Sarah is reading a newspaper when she realizes that four numbered pages of one section are missing. If one of the missing pages is numbered 9 and there are 40 pages in the section. What is the sum of the other three missing page numbered?

Because there are 40 pages, the sum of all four numbers is 2(40+1)=82.  Thus, the sum of the other three numbers is 73.

What does that mean? Please explain in more detail.

The newspaper does not work like a book, basically imagine multiple papers stacked on each other, then the whole stack is folded in half. One paper removed would remove "4 pages". I hope this helps.


Assign 1 through 8 to the vertices of the cube.  Then assign each edge as the absolute difference between its two vertices. What is the minimal number of distinct values for the edges?

-interesting_move

Bump.  Don't tell me no one can answer this.

wereallgonnadie wrote:[hide]3???[/hide]

Yes, but why, and can you give an example of a case that works?  Big hint: [hide]Think about placing a number on one vertex - can any of the three adjacent vertices differ from the first vertex by the same amount?[/hide]

Oooh! Thanks Mathfanatic. Here's my example:

[hide]If you draw a cube on paper then the front face's verticies are 1, 2, 3, 4, clockwise, and the back face's are 5, 6, 7, 8, also clockwise.

Then the differences are either 1, 3, or 4.

If a # is placed on a vertex then two of the adjecent ones can have the same difference. For example, if my # is 2 then the adjacent ones can be 1, 3, and 4. 1 and 3 have the same difference with 2.

But, for 1 you can't have two of the same difference because that would mean the two adjecent numbers are equal! So the least possible amount is 3.[/hide]

By the way, my first one was a lucky guess.


function
integration
real analysis
real analysis unsolved

Find all functions $ f$ satisfy \[ f(x) \plus{} f'(x) \equal{} x^4 \plus{} 3x^3 \minus{} x^2 \plus{} 9x \plus{} 12.\]

An alternate way to obtain a particular solution for this nonhomogeneous linear differential equation - the method of undetermined coefficients:

Assume that $ f(x)\equal{}Ax^4\plus{}Bx^3\plus{}Cx^2\plus{}Dx\plus{}E.$ Just put this into the original equation to get that:

$ A\equal{}1$

$ 4A\plus{}B\equal{}3,$ so $ B\equal{}\minus{}1$

$ 3B\plus{}C\equal{}\minus{}1,$ so $ C\equal{}2$

$ 2C\plus{}D\equal{}9,$ so $ D\equal{}5$

$ D\plus{}E\equal{}12,$ so $ E\equal{}7$

That gives us $ x^4\minus{}x^3\plus{}2x^2\plus{}5x\plus{}7.$ For completeness, we must add an arbitrary solution of the corresponding homogeneous equation, hence the general solution is
\[ f(x)\equal{}x^4\minus{}x^3\plus{}2x^2\plus{}5x\plus{}7\plus{}Ce^{\minus{}x}\]
I would add that this should be in complete agreement with what Kirill said - or would be, had be bothered to evaluate his integral.

Another approach is by means of D-operators

$ f(x) \plus{} f^{\prime }(x) \equal{} x^{4} \plus{} 3x^{3} \minus{} x^{2} \plus{} 9x \plus{} 12$

$ f(x) \equal{} \frac {1}{1 \plus{} D}\left( x^{4} \plus{} 3x^{3} \minus{} x^{2} \plus{} 9x \plus{} 12\right)$

$ \equal{} \left( 1 \minus{} D \plus{} D^{2} \minus{} D^{3} \plus{} D^{4}\right) \left( x^{4} \plus{} 3x^{3} \minus{} x^{2} \plus{} 9x \plus{} 12\right)$

$ \equal{} \left( x^{4} \plus{} 3x^{3} \minus{} x^{2} \plus{} 9x \plus{} 12\right) \minus{} \left( 4x^{3} \plus{} 9x^{2} \minus{} 2x \plus{} 9\right) \plus{} \left( 12x^{2} \plus{} 18x \minus{} 2\right) \minus{} (24x \plus{} 18) \plus{} 24$

$ \equal{} \allowbreak x^{4} \minus{} x^{3} \plus{} 2x^{2} \plus{} 5x \plus{} 7$

Adding on the homogeneous solution $ Ce^{ \minus{} x},$

we have  $ f(x) \equal{} \allowbreak x^{4} \minus{} x^{3} \plus{} 2x^{2} \plus{} 5x \plus{} 7 \plus{} Ce^{ \minus{} x}$


If the line $L$ in the $xy$-plane has half the slope and twice the y-intercept of the line $y = \frac{2}{3} x + 4$, then an equation for $L$ is:

$ \textbf{(A)}\  y = \frac{1}{3} x + 8   \qquad
\textbf{(B)}\  y = \frac{4}{3} x + 2   \qquad
\textbf{(C)}\   y = \frac{1}{3} x + 4  \qquad\\
\textbf{(D)}\   y = \frac{4}{3} x + 4  \qquad
\textbf{(E)}\   y = \frac{1}{3} x + 2   $

[quote="Silverfalcon"]If the line $L$ in the $xy$-plane haas half the slope and twice the y-intercept of the line $y = \frac{2}{3} x + 4$, then an equation for $L$ is:

A. $y = \frac{1}{3} x + 8$
B. $y = \frac{4}{3} x + 2$
C. $y = \frac{1}{3} x + 4$
D. $y = \frac{4}{3} x + 4$
E. $y = \frac{1}{3} x + 2$[/quote]

Ehh...

[hide]
$y = \frac{\frac{2}{3}}{2} x + 2\cdot4$ = $y = \frac{1}{3} x + 8$

(A)?
[/hide]

Are you sure this is an AHSME problem?  :?

[hide]$\frac{1}{2}\cdot\frac{2}{3}=\frac{1}{3}$. Also 4*2=8. The equation is then
$y=\frac{1}{3}x+8$. [/hide]

Quite possibly a #1-3. Remember, everyone is supposed to be able to get [b]some[/b] of the questions right, and this just requires a basic knowledge of Algebra I. ;)

[quote="Silverfalcon"]If the line $L$ in the $xy$-plane haas half the slope and twice the y-intercept of the line $y = \frac{2}{3} x + 4$, then an equation for $L$ is:

A. $y = \frac{1}{3} x + 8$
B. $y = \frac{4}{3} x + 2$
C. $y = \frac{1}{3} x + 4$
D. $y = \frac{4}{3} x + 4$
E. $y = \frac{1}{3} x + 2$[/quote]
[hide]
In the equation $y = \frac{2}{3} x + 4$, $\frac{2}{3}$ is the slope. half the slope is $\frac{\frac{2}{3}}{2}=\frac{1}{3}$.
The $4$ in $y = \frac{2}{3} x + 4$ is the y intercept. Twice this is $4*2=8$.
So our line is $y=\frac{1}{3}x+8$. Our answer is [b]A[/b].
[/hide]

I think this is too easy to be a AHSME problem but,

the answer is A.

[hide]
$\frac{1}{2}\cdot\frac{2}{3}=\frac{1}{3}$ and $4\cdot2=8$
So the answer is $y=\frac{1}{3}x+8$ or $A$[/hide]


a car is travelling on a sine curve accordding to the equation along the $XY$ plane

$y=b\sin(\frac{x}{\alpha})$
the coefficent of static friction is $\mu$ between the road and the car(assume it as a point mass
) tyres and the road .
find the maximum speed $v_{max}$  with which the car can uniformly(speed)  move along the path without sliding?

even i don't know the solution :( .i just know the answer.
 the answer is $v_{max}=\alpha \sqrt{\frac{\mu g}{b}}$

On a horizontal road the only force that provides centripetal acceleration is the friction force:

     $\,\,F\,\le\,\mu.m.g \,.\quad$   So,     $\quad\,\,{mv^{2}\over \rho}\,\le\,\mu.m.g$ where $\rho\,$ represents the curvature radius.

For the given curve  $\, \rho\,={a^{2}\over b}.{{[1+({b \over a})^{2}.cos^{2}B]^{3\over2}}\over{sin B}\,\,}$, where  $\,\,B\,=\,{x \over a}\,\,$, whose only minimum occurs for cos B = 0 .

 This gives  $\rho_{min}\,=\,{a^{2}\over b}$.

Using this in the inequation $\,\,{v^{2}\over\rho}\le\,\mu.g$ follows your answer.

how does one prove for a sine curve trajectory $F_{c}$ $=$ $\frac{mv^{2}}{\rho}$

this is true by def. or by do you have  a proof for it

[quote="Thaakisfox"]If a particle is moving on an arbitrary curve, then his movement can be broken into small bits. so on a small bit the particle is moving on a circular path, on that circle which is closely touching the curve at the given point. The radius of this circle can be calculated at the given point of a known curve. Hence it is true. :D[/quote]
can u be a bit more mathematical in ur explaination

You can find it, for instance, in

http://ocw.mit.edu/NR/rdonlyres/Aeronautics-and-Astronautics/16-07Fall-2004/C7E6143C-F82A-4920-AAE7-297B6A521018/0/d4.pdf


Hi,
I seriously need help with this problem:

How do I determine whether a limit exists ie f(x,y) = r/(2pi -theta)

Many thanks in advance

:mad: Here are 3 reasons why AoPSers do not want to help:

1) Seriously, this forum isn't your "SAVE ME! HELP! OR I'LL FAIL MY COURSE IF YOU DON'T REPLY ON TIME!" AoPS/ML doesn't work that way.

2) Your question doesn't make any sense. If you ask how to determine the existence of a limit in Calc III, I wonder how you passed Calc I and Calc II... :huuh:

3) There is no reason to post the same topic all over the places like an ad. It is ---> :spam:
[url]http://www.mathlinks.ro/viewtopic.php?t=203728[/url]


superior algebra
superior algebra unsolved

Consider the following lines in affine space $ \mathbb{A}^3(\mathbb{R})$:
$ l_1\equal{}${$ x_1\equal{}x_2\equal{}0$}, $ l_2\equal{}${$ x_1\equal{}x_3\equal{}0$}, $ l_3\equal{}${$ x_2\equal{}x_3\equal{}0$}.
Compute generators for $ I(l_1 \cup l_2 \cup l_3)$.

Is it 0?


[quote="perfect_radio"] wrote:

Note 1. One could also try to prove that

$\sum \frac1{a + b^2 + 1} \leqslant 1$
holds in the given conditions. At least, I hope it holds.[/quote]

$\sum \frac1{a + b^2 + 1} \leqslant 1\Leftrightarrow$
$\Leftrightarrow a^3c^2+b^3a^2+c^3b^2+a^3b+b^3c+c^3a\geq a^2+b^2+c^2+a+b+c.$
I think, if the last inequality is true, it is very interesting.
$a^3c^2+b^3a^2+c^3b^2\geq a+b+c$ is obviously true.
But $a^3b+b^3c+c^3a\geq a^2+b^2+c^2$ is wrong.  ;)

[quote="perfect_radio"]
Actually, I wanted to ask you something:

[quote="arqady"]But $\sum_{cyc}\frac{1}{ab+b+1}=1$.[/quote]

How did you see this? When faced with it, it's not hard to prove, but how did you come up with it from thin air?[/quote]
This is very very old and known problem. :)

$a^3c^2+b^3a^2+c^3b^2+a^3b+b^3c+c^3a\geq a^2+b^2+c^2+a+b+c,$
where $a,$ $b$ and $c$ are positive, such that $abc=1.$
I think it is true, but my proof is  very ugly.

[quote="perfect_radio"]Let $a,b,c > 0$ s.t. $abc=1$. Prove that \[ \sum \frac1{a^2 + 2 b^2 + 3} \leqslant \frac12 .  \]

[i]Mircea Lascu[/i]

[b]Note 1.[/b] One could also try to prove that \[ \sum \frac1{a + b^2 + 1} \leqslant 1  \] holds in the given conditions ($abc=1$). At least, I hope it holds.

[b]Note 2.[/b] I hope that the inequality hasn't been discussed until now. I entered very rarely in this sub-forum in the last months  :blush: (and, yeah, searching turned up nothing)[/quote]
I think the first ineq is composed by two inequality:
$\sum_{cyc}{\frac{1}{a^2+b^2+1}}\leq 1$
$\sum_{cyc}{\frac{1}{a^2+2}\leq 1}$
both these two can be solved easily,and I have seen them in the forum.

arqady, is your solution similar to [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=566345#566345]maky's solution[/url]?

[quote="Sung-yoon Kim"][quote="perfect_radio"]Indeed, that's what I found (for the second one, I mean), i.e. I found nothing.

Actually, I wanted to ask you something:

[quote="arqady"]But $\sum_{cyc}\frac{1}{ab+b+1}=1$.[/quote]

How did you see this? When faced with it, it's not hard to prove, but how did you come up with it from thin air?[/quote]

As $abc=1$, we can substitute like $a=\frac{p}{q}, b=\frac{q}{r}, c=\frac{r}{p}$. Then $\sum_{cyc}\frac{1}{ab+b+1}=\sum_{cyc}\frac{r}{p+1+r}=1$.[/quote]

It should be $\sum_{cyc}\frac{1}{ab+b+1}=\sum_{cyc}\frac{r}{p+q+r}=1$.

[quote="perfect_radio"]
[b]Note 1.[/b] One could also try to prove that \[\sum \frac1{a+b^{2}+1}\leqslant 1\] holds in the given conditions ($abc=1$). At least, I hope it holds.

[/quote]

Yes, this holds.

[quote="arqady"]$\sum \frac1{a+b^{2}+1}\leqslant 1\Leftrightarrow \Leftrightarrow a^{3}c^{2}+b^{3}a^{2}+c^{3}b^{2}+a^{3}b+b^{3}c+c^{3}a\geq a^{2}+b^{2}+c^{2}+a+b+c$.[/quote]

From here, as $abc=1$, we can put $a=\frac{q}{p},b=\frac{r}{q},c=\frac{p}{r}$.
Then the inequality is equivalent to
$\sum_{cyc}p^{5}q^{4}+\sum_{cyc}p^{6}qr^{2}\geq \sum_{cyc}p^{4}q^{3}r^{2}+\sum_{cyc}p^{5}q^{3}r$.

Now we use the following inequalities:
$9\sum_{cyc}p^{5}q^{4}+5\sum_{cyc}p^{6}qr^{2}\geq14\sum_{cyc}p^{5}q^{3}r$
from $8p^{5}q^{4}+q^{5}r^{4}+5p^{6}qr^{2}\geq 14p^{5}q^{3}r$,
$6\sum_{cyc}p^{6}qr^{2}+\sum_{cyc}p^{5}q^{4}\geq7\sum_{cyc}p^{4}q^{3}r^{2}$
from $3p^{6}qr^{2}+2p^{2}q^{6}r+pq^{2}r^{6}+p^{5}q^{4}\geq7p^{4}q^{3}r^{2}$, and
$\sum_{cyc}p^{5}q^{4}\geq\sum_{cyc}p^{4}q^{3}r^{2}$
from $4p^{5}q^{4}+q^{5}r^{4}+2r^{5}p^{4}\geq7p^{4}q^{3}r^{2}$. (let's number hte inqualities as $(*),(**),(***)$.)

$(*)\times 2+(**)\times 3+(***)\times 7$ gives
$\sum_{cyc}p^{5}q^{4}+\sum_{cyc}p^{6}qr^{2}\geq \sum_{cyc}p^{4}q^{3}r^{2}+\sum_{cyc}p^{5}q^{3}r$
and this completes the proof.

I know this is really old but I have a question:

Is there any kind of method (like the muirhead method) by which Kim found the result in his last post? Did it involve solving systems of equations etc.?

Any discussion would be appreciated :)

edit: here's what I just found:

[quote="perfect_radio"]E bine :D

Se pare că sistemul care se obţine are trei ecuaţii şi şase necunoscute:

$\left\{ \begin{array}[c]{c}0x-5y-1z+1u-3v-4v=0 ; \\ 3x+4y-z+0u+5v+1w=0 ; \\-3x+1y+2z-1u-2v+3w=0 . \end{array}\right.$

Spatiul $\textrm{ker}\, A$, unde $A = \begin{pmatrix}0 &-5 &-1 & 1 &-3 &-4 \\ 3 & 4 &-1 & 0 & 5 & 1 \\-3 & 1 & 2 &-1 &-2 & 3 \end{pmatrix}$, este "spawned" de
\[\left\{ \begin{pmatrix}11 \\-12 \\ 0 \\ 0 \\ 0 \\ 15 \end{pmatrix}, \begin{pmatrix}-13 \\-9 \\ 0 \\ 0 \\ 15 \\ 0 \end{pmatrix}, \begin{pmatrix}-4 \\ 3 \\ 0 \\ 15 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix}3 \\-1 \\ 5 \\ 0 \\ 0 \\ 0 \end{pmatrix}\right\}. \]
$\begin{pmatrix}1 \\ 10 \\ 103 \\ 223 \\ 10 \\ 10 \end{pmatrix}$ aparţine $\textrm{ker}\, A$ deoarece
\[\begin{pmatrix}1 \\ 10 \\ 103 \\ 223 \\ 10 \\ 10 \end{pmatrix}= \frac23 \cdot \begin{pmatrix}11 \\-12 \\ 0 \\ 0 \\ 0 \\ 15 \end{pmatrix}+\frac23 \cdot \begin{pmatrix}-13 \\-9 \\ 0 \\ 0 \\ 15 \\ 0 \end{pmatrix}+\frac{223}{15}\cdot \begin{pmatrix}-4 \\ 3 \\ 0 \\ 15 \\ 0 \\ 0 \end{pmatrix}+\frac{103}5 \cdot \begin{pmatrix}3 \\-1 \\ 5 \\ 0 \\ 0 \\ 0 \end{pmatrix}. \]
De asemenea, programul cu care am făcut calculele de mai sus mi-a dat următoarea soluţie:
\[\begin{pmatrix}1 \\ 1 \\ 13 \\ 25 \\ 1 \\ 1 \end{pmatrix}= \frac1{15}\cdot \begin{pmatrix}11 \\-12 \\ 0 \\ 0 \\ 0 \\ 15 \end{pmatrix}+\frac1{15}\cdot \begin{pmatrix}-13 \\-9 \\ 0 \\ 0 \\ 15 \\ 0 \end{pmatrix}+\frac53 \cdot \begin{pmatrix}-4 \\ 3 \\ 0 \\ 15 \\ 0 \\ 0 \end{pmatrix}+\frac{13}5 \cdot \begin{pmatrix}3 \\-1 \\ 5 \\ 0 \\ 0 \\ 0 \end{pmatrix}, \]
adică $15 s_{1}+27 s_{2}\geq 42 t_{2}$.[/quote]

Can anyone explain (esp. perfect radio :o )

[quote="arqady"]$\sum \frac1{a+b^{2}+1}\leqslant 1\Leftrightarrow$
$\Leftrightarrow a^{3}c^{2}+b^{3}a^{2}+c^{3}b^{2}+a^{3}b+b^{3}c+c^{3}a\geq a^{2}+b^{2}+c^{2}+a+b+c.$[/quote]
[quote="maky"]$\Leftrightarrow \sum_\textrm{cyc}x^{6}y z^{2}+\sum_\textrm{cyc}x^{5}y^{4}\geq \sum_\textrm{cyc}x^{4}y^{2}z^{3}+\sum_\textrm{cyc}x^{5}y z^{3}. \ \left( \text{Letting}\ a = \frac{x}{y}, \, b = \frac{y}{z}, \, c = \frac{z}{x}.\right)$[/quote]
[quote="maky"]$s_{1}= \sum_\text{cyc}x^{6}y z^{2}$
$s_{2}= \sum_\text{cyc}x^{5}y^{4}$
$t_{1}= \sum_\text{cyc}x^{4}y^{2}z^{3}$
$t_{2}= \sum_\text{cyc}x^{5}y z^{3}$
Summing cyclically the inequalities
$8x^{5}y^{4}+2y^{5}z^{4}+11z^{5}x^{4}\geq 21x^{4}y^{2}z^{3},$
and
$4x^{6}yz^{2}+y^{6}zx^{2}+2z^{6}xy^{2}\geq 7x^{4}y^{2}z^{3},$
we get that
$s_{1}\geq t_{1}$ and that $s_{2}\geq t_{1}.$
Hence, $\alpha s_{1}+\beta s_{2}\geq \left( \alpha+\beta \right) t_{1}$, for any $\alpha,\beta > 0$.
We want to find $\alpha, \beta > 0$ such that $\beta s_{1}+\alpha s_{2}\geq \left( \alpha+\beta \right) t_{2}$.
[/quote]
By AM-GM,
$a x^{6}y z^{2}+b x^{2}y^{6}z+c x y^{2}z^{6}+u x^{5}y^{4}+v y^{5}z^{4}+w z^{5}x^{4}$
$\geq (a+b+c+u+v+w) \left( x^{6a+2b+c+5u+4w}\right)^{1/(a+b+c+u+v+w)}y^{\ldots}z^{\ldots}.$
You want the $RHS$ to be equal to $x^{5}y z^{3}$. Now just solve the system.


geometry
geometric transformation
homothety

if r[a] is the radius of the excircle tangent to side a of triangle ABC, and s is the semiperimeter of ABC, prove:

r[b]*r[c]+r[a]*r[c]+r[a]*r[b] = s2

Geometry is a wonderful thing.

[hide]We have easily that r[a]*(b + c - a)/2 = K, where K is the area of the triangle.  Thus, r[a] = K/(s - a).

Plug in:

r[a]*r[b] = K2/((s - a)(s - b)) = s*(s - c) by Heron's.   Similarly

r[b]*r[c] = K2/((s - b)(s - c)) = s*(s - a)

r[c]*r[a] = K2/((s - c)(s - a)) = s*(s - b)

Adding, we get r[a]r[b] + r[b]r[c] + r[c]r[a] = s(3s - a - b - c) = s2[/hide]

JBL wrote:[hide]We have easily that r[a]*(b + c - a)/2 = K, where K is the area of the triangle.[/hide]

I don't think this is a trivial assertion. You should justify this.

I prefer to not spam people with pictures unnecessarily. ^_^

Yeah, that extra .06 seconds of loading into your browser, hehe. Cheers! :D

** uses [b]bastard[/b] logic **

Suppose that 100 people views this thread, so that'd be 60 seconds! ONE FULL MINUTE! YOU'VE JUST WASTED ONE FULL MINUTE! It could have been used to do something productive, like to kill a fly!

... anyhow, that's a really neat result.

(hehehe... yes, the mistake is on purpose)

JBL wrote:By areas -- [hide]if O[a] is the excenter opposite A, then we have that the area of quadrilateral CABO[a] is the area of ABC plus the area of BCO[a] and the area of ABO[a] plus the area of ACO[a].  The heights of the three triangles with O[a] as a vertex are all r[a], and their bases are a, b, and c.  Thus, we get the stated result.[/hide]

I used a different way of getting this lemma:

the length of the tangent from  vertex A to the opposing excircle is s (semiperimeter), and the length of the tangent from that vertex to the incircle is (b+c-a)/2. Therefore, by homothety, r*s = r[a] * (b+c-a)/2. But r*s = K, so K = r[a]*(b+c-a)/2.

But then again, I just like homothety.

By homothety, do you mean similar triangles?

most of homothety is proven through similar triangles, so essentially it is. i just like the name


calculus
integration
logarithms
Putnam
real analysis
real analysis unsolved

Let $ a>0$ be a positive real number. Calculate:

1) $ \int\limits_{0}^{a}\frac{\ln(x\plus{}a)}{x^{2}\plus{}a^{2}}dx$,

and

2) $ \int\limits_{a}^{2a}\frac{\ln x}{x^{2}\plus{}2ax\plus{}2a^{2}}dx$.

"Riemann" is irrelevant here- these definite integrals make sense in any definition.

The first is only an easy substitution away from [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=64436]Putnam 2005 A5[/url].

[quote="vector2"][quote]The first is only an easy substitution away from Putnam 2005 A5. [/quote]

Would there be a text or paper that discusses this solution method?

I'm interested and think it would be a valuable read?

thanks[/quote]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=383281#p383281]Here[/url].

That's a redundant link, sylow_theory.

[url=http://en.wikipedia.org/wiki/Differentiation_under_the_integral_sign]This[/url] is the method I used- but I prefer writing everything in integral language, where I know when the interchange is justified.
There's a fundamental principle of mathematics: the reason to write down an iterated sum or integral is that we intend to switch the order.

[quote="jmerry"]That's a redundant link, sylow_theory.[/quote]
I really am sorry, I didn't see it. I can't delete my post now, :(

EDIT: For #2 woudn't $ t\equal{}x\plus{}a$ do the integral, because $ x^2\plus{}2ax\plus{}2a^2\equal{} t^2 \plus{}a^2$ and it almost has the same form as #1 so maybe the trick is gonna work.

It's a different numerator, and a different interval. Similar techniques might work, but I haven't looked closely.


Two operations are allowed:
(i) To write two copies of number 1.
(ii) To replace any two identical numbers $n$ by $n+1$ and $n-1$.

Find the minimal number of operations required to produce the number 2005. (At the beginning there are no numbers)

This is problem 7 of the tournament of the cities :)

The key is to observe that, if

$F=\sum p^2$ where p is any number written on the board,
$F$ grows with two on each turn.

The second hook is to prove that in the best variant the final numbers written on the board are 0,...,0,1,1,2,3,4,...,2003,2005.

The final answer is [b]1342355520[/b] :)

The problem is very similar to the problem of the 'fleas'.

[quote="imortal"]

The problem is very similar to the problem of the 'fleas'.[/quote]

Which one?

This one :) : http://www.mathlinks.ro/Forum/viewtopic.php?t=74658


quadratics
modular arithmetic
number theory unsolved
number theory

Let positive integers $ x,y,n$ such that $ n\ge 2,x\ge 3, x^2\plus{}5\equal{}y^n$. Suppose $ p$ is an arbitrary prime divisor of $ n$. Prove that $ 4|(p\minus{}1)$

[quote="thaithuan_GC"]Let positive integers $ x,y,n$ such that $ n\ge 2,x\ge 3, x^2 + 5 = y^n$. Suppose $ p$ is an arbitrary prime divisor of $ n$. Prove that $ 4|(p - 1)$[/quote]
If n is an even integer then it is quite easy to check  it doesn't satisfy condition .
Now consider n is an odd number .Call $ p$ is a prime divisor of $ n$ . Let $ a = y^{\frac {n}}{p}$ then we have :
$ x^2 + 5 = a^p$
It is easy to check that $ a$ is an odd number (if not we will have $ 8|x^2 + 5$,it gives contradiction ) ,therefore x is an even number .
Suppose that $ p$ has form $ 4k + 3$ . From the equation we have :
\[ x^2 + 4 = a^p - 1 = (a - 1)(\frac {a^p - 1}{a - 1})
\]
Because $ \frac {a^p - 1}{a - 1}\equiv 1 (\mod 2)$ and $ 4|x^2 + 4$ so $ 4|a - 1$ . It gives that :
$ \frac {a^p - 1}{a - 1}\equiv p\equiv 3 (\mod 4)$ . It means that this number has at least prime divisor of the form $ 4k + 3$ ,but it is not suitable with the fact that all odd prime divisor of $ x^2 + 4$ has form 4t+1.
So $ p$ is has form $ 4k + 1$ .
Proof completes .

Could You explain more clearly why $ x^2\plus{}4$ has divisors only in the form $ 4k\plus{}1$?

Because I don't know how to prove this statement but I understand why from this follows the conclusion.

thanks for Your reply

if p divides $ x^2\plus{}4$ we observe that -4 is a quadratic residue mod p.
By legendre symbol we then have $ (\minus{}4/p)\equal{}1$
But we know  $ (\minus{}4/p)\equal{}(4/p)*(\minus{}1/p)$
and we know $ (4/p)$=$ (2/p)*(2/p)$
But $ (2/p)$=$ (\minus{}1)^{p^2\minus{}1/8}$
and $ (\minus{}1/p)\equal{}(\minus{}1)^{p\minus{}1/2}$
implying that $ (\minus{}4/p)$=$ (\minus{}1)^{p^2\plus{}2p\minus{}3/4}$
so we want $ (p^2\plus{}2p\minus{}3)/4$ to be even
putting primes of the form $ 4k\plus{}3$ results in $ (p^2\plus{}2p\minus{}3)/4$ being odd.
which implies all prime divisors are of the form $ 4k\plus{}1$

we can simplify befuddlers argument a little bit- $ (\frac{2}{p})\cdot (\frac{2}{p})\equal{}(\frac{2}{p})^2\equal{}1$ when $ p \neq 2$, so we only need $ \frac{p\minus{}1}{2}$ to be even. Alternatively, $ x^2\plus{}4 \equiv 0 \pmod p \implies (2^{\minus{}1}x)^2 \equiv \minus{}1 \pmod p$, or: $ 1 \equiv (2^{\minus{}1}x)^{p\minus{}1} \equiv (\minus{}1)^{\frac{p\minus{}1}{2}} \pmod p$, and again $ 2 \mid \frac{p\minus{}1}{2}$, or $ 4|p\minus{}1$


I was thinking, maybe the LaTeX on the forum could have an equation checker that checks equations you type. It would warn you if it is not valid. But then a calculator would be needed.

Also, a calculator could be provided to users on the corner of the screen, or they could have a hide/show option.

The correct solution to the first problem is to use the preview option and carefully edit your posts for correctness of LaTeX and correctness of mathematics before posting.  This is an obvious and important part of "making sure your post is meaningful and intelligible."  Every widely-used operating system comes with a built-in calculator.  If you can't check your arithmetic before posting, this is very much your problem, not the problem of the site administrators.


$Three chords  AB , CD , EF  of a circle form angles  60ο when each one intersect the two others. AB and CD meet at K , AB and EF meet at  L and finally  CD  and EF meet at M , so that triangle  KLM  is equilateral .
Prove that
KA + LE + MD = KC + LB + MF $

There was a problem in sending  post(post 1).
Please see the correct  topic :

    http://www.mathlinks.ro/Forum/viewtopic.php?t=51696

This pdf file contains the figure of that  topic.
  Sorry.


How many pairs of positive integers (a,b) are there such that gcd(a,b)= 1 and a/b + 14b/9a is an integer ?


Let $p \ge 5$ be a prime number. Prove that there exists an integer $a$ with $1 \le a \le p-2$ such that neither $a^{p-1} -1$ nor $(a+1)^{p-1} -1$ is divisible by $p^2$.

We will consider 3 cases:
1)$(\frac{p-1}{2})^{p-1} \not \equiv 1 \mod{p^2}$ and $(\frac{p+1}{2})^{p-1} \not \equiv 1\mod{p^2}$
2)$(\frac{p-1}{2})^{p-1} \equiv 1 \mod{p^2}$
3)$(\frac{p+1}{2})^{p-1} \equiv 1 \mod{p^2}$

In the first case, we obviously take $a=\frac{p-1}{2}$.
Let us consider the second case.
$(\frac{p-1}{2})^{p-1} \equiv 1 \mod{p^2}$ is equivalent to $(p-1)^{p-1} \equiv 2^{p-1} \mod{p^2}$ which is, after reducing $\mod{p^2}$,
$p(p-1) + 2^{p-1} - 1 \equiv 0 \mod{p^2}$
From FLT $2^{p-1} - 1$ is divisible by $p$ so, $2^{p-1} - 1 = pk$. So:
$p(p-1) + 2^{p-1} - 1 = p(p-1 + k) \equiv 0 \mod{p^2}$
So, $p-1+k \equiv 0 \mod{p}$ which means that $k \equiv 1 \mod{p}$ and $2^{p-1} - 1 \equiv p \mod{p^2}$

Now, we will show that $a=p-2$ satisfies required condition. Indeed:

$(p-2)^{p-1} - 1 \equiv 2^{p-2}p(p-1) + 2^{p-1} - 1 \equiv p(2^{p-2}(p-1) + 1) \mod {p^2}$
And since $2^{p-2}(p-1) + 1 \equiv -2^{p-2} + 1 \not \equiv 0 \mod{p}$ above congruence is not $0 \mod{p^2}$.

Now we check $p-1$:
$(p-1)^{p-1} - 1 \equiv -p(p-1) + 1 - 1 \equiv -p(p-1) \not \equiv 0 \mod{p^2}$

So, in this case we are done.
We are left with the case 3) but it's almost the same.

Another solution (I suggest):
1) If $p^2|a^{p-1}-1$ then $(p-a)^{p-1}-1$ is not divisible by $p^2$
We have, if $p^2|(p-a)^{p-1}-1=p^{p-1}-\binom{p-1}{1}.p^{p-2}.a+...-\binom{p-1}{p-2}.p.a^{p-2}+a^{p-1}-1$ then $p^2|\binom{p-1}{p-2}p.a^{p-1}$ which is a contradiction
2) IF $p^2|a^{p-1}-1$ and $p^2|(a+1)^{p-1}-1$ then $(p-a)^{p-1}-1$ and $(p-a-1)^{p-1}-1$ aren't divisible by $p^2$ then we finish the problem because we choose $a'=p-a-1$ then $a'^{p-1}-1$ and $(a'+1)^{p-1}-1$ are not divisible by $p^2$
3) IF $a^{p-1}-1,(a+1)^{p-1}-1$ there is only one number which is divisible by $p^2$ $(1)$ then we have:

i) We have if $p^2|2^{p-1}-1$ then by the statement above, we have $(p-2)^{p-1}-1$ is not divisible by $p^2$
So we choose $a=p-2$ then $(p-2)^{p-1}-1$ is not divisible by $p^2$ and since $p^2|1^{p-1}-1 $ then $(p-1)^{p-1}-1$ is not divisible by $2$ so $a=p-2$ satisfies the problem
ii) If $2^{p-1}-1$ is not divisible by $p^2$ then by $(1)$ we have $p^2|3^{p-1}-1$ and $4^{p-1}-1$ is not divisible by $p^2$, .... then $(2k)^{p-1}-1$ is not divisible by $p^2$ and $p^2|(2r+1)^{p-1}-1$ for all $k,r$ $(2)$
We consider the number $(2k)^{p-1}-1$ is not divisible by $p^2$ and $p^2|(p-2k)^{p-1}-1$ with $k$ is an odd number, we have $p^2|(p-2k)^{p-1}-1 \Rightarrow p^2|(2k)^{p-1}-(p-1)p(2k)^{p-2}-1$
$\Rightarrow p^2|(2k)^{p-1}+p(2k)^{p-2}-1$ and since $k$ is odd then $p^2|k^{p-1}-1$ by $(2)$
Thus $p^2|k.2^{p-1}+p2^{p-2}.k^{p-1}-k$
$ \Rightarrow p^2|k.2^{p-1}+p.2^{p-1}-k$
$ \Rightarrow k.(2^{p-1}-1) \equiv 1-p.2^{p-1} \pmod{p^2}$
But $p=5$ the problem is trivial, and $p>5$ we have there is an other odd number $k'$ which $k'\neq k$, similarly, we gain $k'.(2^{p-1}-1) \equiv 1-p.2^{p-1} \pmod{p^2}$
Thus $k.(2^{p-1}-1) \equiv k'.(2^{p-1}-1) \equiv 1-p.2^{p-1} \pmod{p^2}$ then $p^2|(k'-k)(2^{p-1}-1)$ which is clearly impossible because $k,k'<p$ and $p$ is an odd prime and we are considering $2^{p-1}-1$ is not divisible by $p^2$
Then we finished our problem

Another solution : Case $p=5$ is trivial. So let's consider case $p>5$

Let $S=\{k\in\mathbb{N}\mid 1\le k\le p-1, p^2\nmid k^{p-1}-1\}$ thus we need to show that set $S$ contain 2 consecutive number in it.

First assume that both $a, p-a\notin S$ $a\in\mathbb{N}$ and $1\le a\le p-1$ so $p^2\mid a^{p-1}-1$ and $p^2\mid (p-a)^{p-1}-1$
But binomial theorem gives us that $\displaystyle{(p-a)^{p-1}\equiv\binom{p-1}{1}pa^{p-2}+1\equiv -pa^{p-2}+1\pmod{p^2}}$
So $p^2\mid pa^{p-2}$ which means $p\mid a$ contradiction!

Hence $a\in S$ or $p-a\in S$.

It's easy to show that $1\notin S$ and $p-1\in S$

Assume that for every $a\in S$ implies that $a+1\notin S$

By sequences of deduction from what we have proven gives us $S=\{2,4,6,...,p-1\}$. We'll split into 2 cases.

[hide=Case 1 : 4|p-1]

So $\dfrac{p+1}{2}$ is odd.

But $1\le\dfrac{p+1}{2}\le p-1$ so $\left(\dfrac{p+1}{2}\right)^{p-1}\equiv 1\pmod{p^2}\Rightarrow (p+1)^{p-1}\equiv 2^{p-1}\pmod{p^2}$

Applying binomial theorem gives us $\displaystyle{2^{p-1}\equiv\binom{p-1}{1}p+1\equiv 1-p\pmod{p^2}}$

On the other hand, $\dfrac{p+5}{2}$ is odd and $1\le\dfrac{p+5}{2}\le{p-1}$ because $p>5$

Thus $\left(\dfrac{p+5}{2}\right)^{p-1}\equiv 1\pmod{p^2}$ so $(p+5)^{p-1}\equiv 2^{p-1}\pmod{p^2}$

Applying binomial theorem again gives $\displaystyle{\binom{p-1}{1}p5^{p-2}+1\equiv 1-p\pmod{p^2}\Rightarrow p(5^{p-2}-1)\equiv 0\pmod{p^2}}$

Thus $5^{p-2}\equiv 1\pmod p\Rightarrow 5^{p-1}\equiv 5\pmod p\Rightarrow 1\equiv 5\pmod p$ Contradiction!
[/hide]

[hide=Case 2 : 4|p-3]
So $\dfrac{p-1}{2}$ is odd.

But $1\le\dfrac{p-1}{2}\le p-1$ so $\left(\dfrac{p-1}{2}\right)^{p-1}\equiv 1\pmod{p^2}\Rightarrow (p-1)^{p-1}\equiv 2^{p-1}\pmod{p^2}$

Applying binomial theorem gives us $\displaystyle{2^{p-1}\equiv\binom{p-1}{1}p(-1)^{p-2}+1\equiv 1+p\pmod{p^2}}$

On the other hand, $\dfrac{p+3}{2}$ is odd and $1\le\dfrac{p+3}{2}\le{p-1}$ because $p>5$

Thus $\left(\dfrac{p+3}{2}\right)^{p-1}\equiv 1\pmod{p^2}$ so $(p+3)^{p-1}\equiv 2^{p-1}\pmod{p^2}$

Applying binomial theorem again gives $\displaystyle{\binom{p-1}{1}p3^{p-2}+1\equiv 1+p\pmod{p^2}\Rightarrow p(5^{p-2}+1)\equiv 0\pmod{p^2}}$

Thus $3^{p-2}\equiv -1\pmod p\Rightarrow 3^{p-1}\equiv -3\pmod p\Rightarrow 1\equiv -3\pmod p$ Contradiction!
[/hide]


probability
topology
real analysis
real analysis unsolved

let $E$ be a metric space such that:
1/ there exists a dense countable subset of $E$
2/ $E$ is complete

Suppose that $\mu$ is a borel measure on $E$ such that $\mu(E) < \infty$.
Prove that for all $\epsilon > 0$, there exists a compact $K$ satisfying $\mu(K) > \mu(E) - \epsilon$

Fix some $0<M<\mu(E)$. We need to show that there is a compact $K\subset E$ with $\mu(K)\ge M$.

Consider a countable basis of $E$ consisting of open balls. First cover $E$ with basis elements of radius $\le 1$, and let $A_1$ be the union of finitely many of these basis elements s.t. $\mu(A_1)>M$ (this is clearly possible in the given hypotheses). Now cover $A_1$ with basis elements of radius $\le\frac 12$, and let $A_2$ be a union of finitely many of these s.t. $\mu(A_2)>M$. We continue this process, constructing a descending sequence of sets, $(A_n)_n$, s.t. $A_n$ is a union of basis elements of radius $\le\frac 1n$, and $\mu(A_n)>M,\ \forall n$.

Let $K=\overline{\bigcap_nA_n}$. $K$ is a closed and thus complete subset of $E$. $\bigcap_nA_n$ is totally bounded, and, since in every metric space the closure of a totally bounded set is totally bounded, $K$ is also totally bounded, meaning that $K$ must be compact. Finally, $\mu(K)$ is the limit of the decreasing sequence $(\mu(A_n))_n$, so it must be $\ge M$, as desired.

Prove the same thing for any Borel subset of $E$. More precisely:

If $E$ is as above (a Polish space), $A$ is a Borel subset of $E$ and $\mu$ is a Borel probability measure on $E$, then $\mu(A)=\sup_{K}\mu(K)$, the supremum being taken after all compact subsets $K$ of $A$.


A collection of subsets of $ \{1, 2,..., n\}$ has the property that each pair
of subsets has at least one element in common. Prove that there are at
most $ 2^{n\minus{}1}$ subsets in the collection.

[hide="Solution"]After a shared element is designated, there are $ \tbinom{n \minus{} 1}{0} \plus{} \tbinom{n \minus{} 1}{1} \plus{} \ldots \plus{} \tbinom{n \minus{} 1}{n \minus{} 1} \equal{} 2^{n \minus{} 1}$ possible combinations of the remaining elements.[/hide]

What lets you to choose one shared element?

For example set $ \{1,2,3\}$ could have subsets $ \{1,2\}, \{1,3\}, \{2,3\}, \{1,2,3\}$.

[quote="N.T.TUAN"]A collection of subsets of $ \{1, 2,..., n\}$ has the property that each pair
of subsets has at least one element in common. Prove that there are at
most $ 2^{n \minus{} 1}$ subsets in the collection.[/quote]

[hide="Possibility...?"]For odd $ n$, it's $ \binom{n}{n} \plus{} \binom{n}{n \minus{} 1} \plus{} \ldots \plus{} \binom{n}{\lceil{\frac {n}{2}\rceil}} \equal{} \frac {2^n}{2} \equal{} 2^{n \minus{} 1}$.

Each subset with more than $ \frac {n}{2}$ elements has a complement subset with less than $ \frac {n}{2}$ elements. So then (I think) we should choose the subset with more elements because they will always satisfy the property if we choose all subsets with more than $ \frac {n}{2}$ elements.

For even $ n$, it's $ \binom{n}{n} \plus{} \binom{n}{n \minus{} 1} \plus{} \ldots \plus{} \frac {\binom{n}{\frac {n}{2}}}{2} \equal{} \frac {2^n}{2} \equal{} 2^{n \minus{} 1}$.

It's more or less the same thing here, except when you have a subset with $ \frac {n}{2}$ elements, each subset has a complement that also has $ \frac {n}{2}$ elements, so we can only choose half of the total number of subsets with $ \frac {n}{2}$ elements.[/hide]

It is late where I live, thus I hope my solution does not contain holes.

[u]Solution.[/u] Let $ S$ be the set of all subsets, and $ T \equal{} \{(x_1,\hdots, x_n)|x_i\in\{0,1\}\}$.  Consider the function $ f: S\rightarrow T$, where it sends each set $ U\in S$ to an n-tuple $ (u_1,\hdots, u_n)$, where $ u_i \equal{} 1$ if $ i\in U$ and $ u_i \equal{} 0$ if $ i\not\in U$.

If $ A_i,A_j\in S$, then $ f(A_i) \equal{} f(A_j)\iff A_i \equal{} A_j$, that is, $ f$ is injective.  Thus $ |S|\leq |T|$.  By the given condition, every pair of n-tuples has at least two 1s in the same position.  Define the complement of an n-tuple is this case as $ (x_1,\hdots, x_n)^C \equal{} (1,\hdots, 1) \minus{} (x_1,\hdots, x_n)$ modulo 2.  Hence, we have to eleminate all complements of the n-tuples.  Since the set of complements of n-tuples mapped to the set of n-tuples is bijective, at least half of the number of elements of $ T$ do not have preimages in $ S$.  Hence, $ |S| \equal{} |f(S)|\leq |T|/2 \equal{} 2^{n \minus{} 1}$. Clearly,  $ |S| \equal{} 2^{n \minus{} 1}$ can be achieved. $ \Box$

Using recurrence relation is also possible.

[hide="Quick"]Divide the subsets into $ 2^{n\minus{}1}$ families each containing two sets: one set and its complement with respect to $ \{1,2,\ldots ,n\}$. You can pick at most one element from each family, so you're done.[/hide]
Similar to what chinesechess did.

suppose $T={A_1,A_2,...,A_k}$ is the collection with the most elements and so if u add any set $Q$,that is different from all elements of $T$, in $T$ as another element then it must not satisfy the condition i.e. there must be $i\in {1,2,...,k}$ such that $A_i$ and $Q$ are disjoint. Now because a set with $n-1$ elements has $2^{n-1}$ subsets, if $k>2^{n-1}$ then we must have $j_m\in {1,2,...,k}$ such that $m\in A_j$ for every $m=1,2,...,n$ so a contradiction and we're done


Let $f: (0,1)\to\mathbb{R}$ , $f(x)=x^x$. Prove that $f(x)+f(1-x)\geq\sqrt2.$ ;)

f is convex, so it's true by the definition of convex function.


Suppose that $ a$ and $ b$ are are nonzero real numbers, and that the equation $ x^2\plus{}ax\plus{}b\equal{}0$ has solutions $ a$ and $ b$. Then the pair $ (a,b)$ is

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

Brute force is probably the quickest for this one

$ x^{2} \plus{} x \minus{} 2 \equal{} 0$
$ (x \minus{} 1)(x \plus{} 2) \equal{} 0$
$ x \equal{} 1, \minus{}2$
$ a \equal{} 1$
$ b \equal{} \minus{}2$
Choice C

Using Vieta's:
$ ab\equal{}b\\
a\equal{}1\\ \\
\minus{}a\equal{}a\plus{}b\\
\minus{}2a\equal{}b\\
\minus{}2\equal{}b$.
The answer is $ (1, \minus{}2)\, \boxed C$.

also look for a positive discriminant b^2-4ac>0. This leaves only choices c and d. Then it's even easier.

sorry for asking this in 2018, but doesn't the question say " positive solutions a and b"? or is that just bad wording lol

[quote=whatRthose]He probably felt proud of himself for writing this problem and wanted it to look pro so he put that, tho they arent rly positive[/quote]
This is an old AMC problem...

Looks like the actual PDF - which can be found [url="https://sites.google.com/site/mathjunk/amc10"]here[/url] (2002 AMC 10B #10) - did not state the solutions were positive.  Fixed; thanks for pointing it out.

[quote=Isogonal][quote=whatRthose]He probably felt proud of himself for writing this problem and wanted it to look pro so he put that, tho they arent rly positive[/quote]

Cynical much?[/quote]

Dont worry, ive had this feeling like 500 times when i wrote my own problems, and i screwed up too.

Sorry @above
If its an amc then he must be bad at copying


given $ \phi : = \frac{x-1}{x^2+2x}$, does it exist $ W$ neighbourhood of $ \{+\infty\}$ such that $ \phi(W\setminus\{+\infty\})\subseteq(\mathbb{R}\setminus\{0\})$?

in other words, why, in order to get the taylor expansion of $ e^\phi$, we do have first to get the expansion of $ \phi$, and we cannot just substitute $ \phi$ in the expansion of $ e^x$?

phi is the funcition I defined equal to (x-1)/(x^2+2x).


Let $ ABC$ be a triangle. The incircle of $ ABC$ touches $ BC$ at $ A'$ and $ AA'$ meets the incircle again at $ P$. Let $ CP,BP$ meet the incircle of $ ABC$ again at $ N,M$ respectively. Prove that $ AA',BN,CM$ are concurrent.

http://www.mathlinks.ro/Forum/viewtopic.php?t=2619

  darij


geometry
3D geometry
combinatorial geometry
geometry proposed

we have cube which the length of  its edges is $a$; we drop it on the chessboard,prove that this cube can't cover more than $(a+1)^2$ vertices in chessboard.

You mean that the cube is on one of its face on the chessboard? The chessboard does not cut the cube, that's it?
In other words : You ask to prove that we cannot cover more than $(a+1)^2$ unit square with a square of side length $a$, am I right?

Pierre.

It seems like that to me too, but in this case why use a cube, and not just a square?

In that case, it is a direct consequence of Pick's formula, which states that if $P$ is a simple (non self-intersecting) polygon whose vertices are lattice points then :
$[P] = \frac {1} {2} b + i - 1$, where $b$ (resp. $i$) denotes the number of lattice points on the boundary (resp. the interior) of $P$, and $[.]$ denotes area.

Let $S$ be the square corresponding to the side of the cube which lie on the chessboard. Let $C$ be the convex hull of the lattice points covered by $S$. Then $C$ is contained in $S$ so that, from Pick's formula :
$ \frac {1} {2} b + i - 1 = [C] \leq [S] = a^2$.

Using the convexity of $C$, it follows that $p(C) \leq p(S)$, where $p(.)$ denotes the perimeter.
That gives $p(C) \leq 4a$.
Moreover, the distance between two distinct lattice point is at least $1$ then there cannot be more than $4a$ lattice points on the boundary of $C$, which leads to $b \leq 4a$.
Thus $b + i = [C] +  \frac {1} {2} b + 1 \leq a^2 + 2a + 1 = (a+1)^2$, and we are done.

Pierre.


What calculator programs have you found most useful esp. in math contests/class?
(TI83/84) If you [u]really[/u] have the inclination, post the program.   ;)

polynomial solving/factoring and the formula to solve a monic cubic equation, which beats doing algebraic/trigonometric manipulations (though it would be too messy to post it all out here)

MAPLE...without a doubt. Its ability to handle rational numbers and algebraic integration is astounding.

Also, I have Super Mario World on my TI calculator. Useless as a maths program, but great fun in maths lessons.


What is the smallest positive integer that cannot be expressed as the sum of eight or fewer cubes of positive integers?

Thanks :!:

Isnt it just 23?

Cuz you can have any number 1-16 (obvious), but if you use 2 8's, you can obtain numbers from 16-22 inclusive but not 23.

So do you just keep trying numbers until you get one that can't be expressed as a sum of 8 or fewer cubes?

In a way..but this is easy...if im right (am I?)


Well i beleive typing chemical equations here is obviously difficult with latex so how about some chemistry typesetting software.
[url=http://www.mathlinks.ro/viewtopic.php?p=929737#929737]here[/url] have a look at my discussion
with stevem .i think it is essential to have [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=mhchem]mhchem[/url] enabled on this forum.please do post in your reviews it is [b][u]important[/u][/b]

Yes, I absolutely agree with pardesi, typing chemical equations with latex is indeed very inconvenient.

What's wrong with the LaTeX?

$ \text{CO}_{2}\plus{}\text{H}_{2}\text{O}\rightleftharpoons\text{H}_{2}\text{CO}_{3}\rightleftharpoons\text{H}^{\plus{}}\plus{}\text{HCO}_{3}^{\minus{}}$

the thing is u need to write very short codes.
 further suppose i want to mention say the temp on the arrow how would i do that?
 just read the book in the web page of mhchem i provided in the first post

As far as I am concern, I feel perfectly confortable with Latex. If you cannot write something in Latex, just describe it in your own words - it will probably give you less work.

my corncern was about drawings,equations with all conditions written on them like on the books...

Then it would be more easy by first make the draw with some chemistry software (personnaly I use the old fashioned ChemWindow) and then to post that image on the forum. Of course, don't forget there is a limit for the size available for you to post images.

yes that's what i am doing now but u can see how poor the picture quality is the small molecules are almost illegible

There is a free program [url=http://www.acdlabs.com/download/chemsk.html]here[/url]. It has 3-D renders and templates to draw compounds.

even i have some of these type softwares but none of them directly give pictures in .GIF /.jpg/.jpeg format does this one give.Also i am running on Ubuntu Linux

Hmm try print screen. Then just crop whatever you want and save it as a JPEG or GIF.

I'm actually writing a chem book right now in LaTeX using the package mhchem (which was mentioned above), which works out very well.  I also found two packages for typesetting molecular structures: lewis (available from [url]http://ctan.tug.org/[/url]) and xymtex ([url]http://homepage3.nifty.com/xymtex/fujitas3/xymtex/indexe.html[/url]).  Haven't had time to play around with these yet, but maybe if they are worthwhile we can ask them to be enabled on the forum.

great work gregx007 is there a way we can show the tests here in the test forum


$a,b,c,d,e,f$ are positive reals such that  $a+b+c+d+e+f=N$ and $ab+ac+bc=de+df+ef$ . If ${N}\mid{abc+def}$ prove that $N$ is composite.

Agreed@wrong :D

http://www.mathlinks.ro/Forum/viewtopic.php?t=90045 is even stronger as $ab+ac+bc=de+df+ef$ is replaced by $ab+ac+bc \equiv de+df+ef \mod N$.

[quote="VO_DANH_KHACH"][quote="Litlle 1000t"]$a,b,c,d,e,f$ are [b]positive reals [/b]such that  $a+b+c+d+e+f=N$ and $ab+ac+bc=de+df+ef$ . If ${N}\mid{abc+def}$ prove that $N$ is composite.[/quote]
You are wrong![/quote]

Thanks, my wrong. This is right:

$a,b,c,d,e,f$ are [b]positive integers [/b]such that  $a+b+c+d+e+f=N$ and $ab+ac+bc=de+df+ef$ . If ${N}\mid{abc+def}$ prove that $N$ is composite


What is acceleration and what is velocity? My book doesn't describe them very well, and I can't see how there can be a constant change in velocity. Also, how can meters per second squared be explained? And what is the difference between velocity and speed?

basically, velocity is a change  in direction, so a change in direction is a change in velocity. Whenever velocity is changing- in either magnitude or direction-the motion is said to be accelerated. The units of velocity are m/s and we know that a=v/t where t is in seconds so the units of acceleration are m/s/s or m/s^2. Does that help?

yes, thank you very much... :D. by the way, how can an object have constant acceleration? gravity doesn't make sense anymore after i learned about it....

A body have constant acceleration if the sum of the forces acting on it is also constant (recall Newton's second law). The acceleration of gravity, $ g$, is the acceleration gained by a body when the only force acting on it is its weight, which is the atractive force exerted by Earth on the body.

I gave this explanation to a professor at CU. I think it will hold. It attempts to give a mathematical demonstration that explains the conundrum of the 'two falling objects'.
[hide="GRAVITY"]
$ F_{g}=G\frac{m_{1}m_{2}}{r^{2}}$

Oh, how [i]well[/i] you will come to know that formula. That's gravitational force, notated by $ F_{g}$. $ m_{1}$ [b]&[/b] $ m_{2}$ are the two objects between whom a force is generated. $ r$ is the distance between the two objects, and $ G$ is a universal constant, at three significant figures, is equal to $ 6.67\times 10^{-11}\frac{Nm^{2}}{kg^{2}}$.

Now, equate that force to the force of a moving object. Why, you might ask? Assume that our object is falling due gravity, accelerating towards The Earth.

$ F = m_{1}a = G\frac{m_{1}m_{2}}{r^{2}}$

It just so happens that the mass of our falling object cancels!
($ M_{E}$ is the mass of The Earth)

$ a=G\frac{M_{E}}{r^{2}}$

Thus demonstrating that the acceleration of an object due to gravity is not proportional to it's own mass, but to the mass towards which it is accelerates.

Now, a constant acceleration due to gravity can be achieved... in an atmosphere. The object's shape will cause a certain amount of wind resistance at a certain speed, that just happens to be equal to the the force of gravity, thus canceling our acceleration: we fall at a constant velocity if we neglect wind buffeting.[/hide]

Here's some help with comprehension on the idea of velocities and rates:
[hide="RATES"]Ratios are probably one of the most useful mathematical expression we use. This is because we can express one measurement as proportional to another, and use it as an indication of 'place name here'.

The most common one is probably $ V$ to us who enjoy physics. Commonly denoted by $ V_{o}$ (initial velocity) or $ V_{f}$ (final velocity), these ratio is gauged by $ \frac{distance}{time}$, with direction supplementing our distance.

Now, I'm sure you've seen a triple-decker fraction in your math class? Something like....
[hide="This?"]$ \cfrac{2}{\cfrac{3}{4}}$
this actually equals $ \frac{8}{3}$, but that's not important.
What is good to know is that this example is what acceleration is (after a fashion).

It is a rate of change within a rate...which measures change.[/hide]

I hope that helps man, that's all I've got![/hide]

[quote="Igiul"]are the two objects between whom a force is generated.[/quote]

Forces are not "generated", they are just a concept used to describe interactions. Perhaps it would be more correct to say "...the two objects between whom there is an interaction described by the force...".


geometry
rectangle
complementary counting

So these problems are from Intro to Counting and Probability.

I am not sure why I am getting these wrong.

The first one says:

4.17) How many different rectangles can be formed by connecting four of the dots in a 4x4 square array of dots such that the sides of the rectangle are parallel to the sides of the grid?

For this, I made the realization that I can start with any of the 16 dots. then I have two possible ways to go. I can go the right or the left, and up and down. So from every dot, there are 3 dots to the left and right total, and 3 up and down total. Thus, I have (16*3*3)/3! giving me 24.

But the book got 36, as we can view the whole gird as having 4 vertical lines and 4 horizontal lines. then there are $ \binom{4}{2}$ for each, thus $ {\binom{4}{2}}^2 \implies {6}^2 \implies 36.$

I am not sure where I am going wrong.

Second:

4.20) The Aviation club at Rocco's school has 12 members (incl. Rocco). They need to choose a 3-person helicopter Committee and a 4-person Glider Committee. Students can serve on either or both committees, but Rocco refuses to serve on both - he will only serve on one or the other (or neither). On how many ways can both committees be chosen?

The book did it by complementary counting - which was very ingenious. But how would/can this problem be solved by casework or something like that?

Thank you for the help.

[quote="limac"]4.17) How many different rectangles can be formed by connecting four of the dots in a 4x4 square array of dots such that the sides of the rectangle are parallel to the sides of the grid?

For this, I made the realization that I can start with any of the 16 dots. then I have two possible ways to go. I can go the right or the left, and up and down. So from every dot, there are 3 dots to the left and right total, and 3 up and down total. Thus, I have (16*3*3)/3! giving me 24.

But the book got 36, as we can view the whole gird as having 4 vertical lines and 4 horizontal lines. then there are $ \binom{4}{2}$ for each, thus $ {\binom{4}{2}}^2 \implies {6}^2 \implies 36.$

I am not sure where I am going wrong. [/quote]

Notice that dividing by $ 3!$ implies that there are three different "rotational" arrangements that lead to the same rectangle, but that isn't actually true, because given two distances, you already know all points, $ 2!2!$ different ways, so using your method, that would have to bee $ \frac{16\cdot3\cdot3}{2!2!}$, which is $ \frac{144}{4}\equal{}\fbox{36}$.

So, dividing by the 3!, I implied to show that there is going to be 3! different labels of the same rectangle. But silly of me to not notice that there are only 4 possible vertices of a rectangle. Thus, ABCD is the same as saying ACBD or DCBA, etc. So, basically, I have 2! ways to arrange the A and D, and 2! ways to arrange the B and C. Since this method applies to all the rectangles formed, I should have divided 16*9 by 4, as you have said. (I probably got disillusioned or something while doing the problem :p)

Thank you very much r15s11z55y89w21.

Now any ideas on the second one?

Well, the cases are already given in the problem; Rocco can serve on one, the other, or neither. Count the ways each of these cases can happen.

Thank you for the reply ThinkFlow.

I tried the casework method before, unsuccessfully as I was making a silly error. Now it is fine.

Thank you.

Actually, I aroused with another problem.

4.22) There are 5 different pairs of gloves, where the right and the left are distinguishable. Select 4 from the 10 gloves.
a) How many ways can you select 2 pairs of gloves?
b) How many ways are there to select 4 such that 2 of the 4 make a pair?

a) This is relatively simple. There are $ \binom{5}{2}.$

b) It is clear that the answer must be less than $ \binom{10}{4},$ which is 210. So I first dealt with the restriction, as in I counted the number of ways to make the pair. Since we have 10 gloves, we can choose any of the 10 gloves and a opposite hand glove for that. So, we have (10*5)/2 as we are counting each pair twice. Then, we have to choose the other two gloves. We have 8 left, thus we can choose any of the 8, and another glove. So we can have (8*7)/2 as we are counting each pair twice. Therefore, the answer should be 25*4*7 = 100*7 = 700.

There is definitely something wrong with the way I did it, but I am not sure what it is though.

Thank you for the help.

[hide]Is the answer by any chance 120? (I want to be sure before I post a solution.)[/hide]

Ihatepie, the correct answer is 210-80 = 130.

Ahh, I think I see my mistake. I thought about the situation as if any two opposite hands glove would make a pair. But that is not the case here according to the Solutions Manual.

Here's another way.

What I did is this:

There are 5 ways to choose a pair. Now, there are 8*7 ways to choose 2 of the remaining 8 gloves. But 4 of these ways involves you getting another pair so there are 56-4 or 52 ways. But this assumes that order is important, which it isn't. So divide it by 2 to get 26. 26*5=130.

Ah ok. You way was what I have been trying to do. But I failed to realize the 56-4 part, and always ended up with 140.

I see it now. Thank you for the help :)

I'm having difficulty understanding one aspect of this problem.

4.23) In poker, a 5-card hand is called a three of a kind is there are three cards of one rank and two other cards which are not the same rank as each other or as the other three cards. How many 5-card hands are three of a kind? (Assume we are using a standard 52-card deck with 13 ranks in each of 4 suits.)

For the solutions, the book did the following:

The three cards of the same rank can be any of the 13 ranks, and the suits have $ \binom{4}{3}$ possible combinations. The next card can be any of the 12 remaining ranks and any of the 4 suits. The card can be any of the 11 remaining ranks and any of the 4 suits. However, the last two cards are interchangeable (their order doesn't matter), so we must divide 2! to correct fro overcounting. Thus the number of 3-of-a-kind hands is $ \frac{(13 * 4)*(12*4)*(11*4)}{2!} \equal{} 54,912.$

The one thing that I don't understand is why are only the fourth and the fifth interchangeable, if we count the first three as a block, aren't all of the three interchangeable and thus shoudl have been divided by 3! instead of 2!? If only the first two are interchangeable why so? And why is the third not?

Thank you for the time to answer these questions.


[quote="sharkman"]March 14 is often celebrated as pi day because the date can be expressed as
3.14, a numerical approximation for the value of p. March 14, 2007 falls on
a Wednesday. At some time in history humans realized that circles appeared
in many sizes and that the greater the distance across a circle the greater
the distance around the circle. There is some evidence that about 2000 BC,
the Babylonians and the Egyptians knew that the ratio between the distance
around a circle (the circumference) and the distance across a circle (the
diameter) is a constant. The symbol “p” was first used by William Jones in
1706 to represent the ratio of a circle’s circumference to its diameter.
The modern representation is C/D = p.

Around 2000 BC the Babylonians approximated the value of p to be 3 1/8
and the Egyptians approximated the value of p to be 4(8/9)2. What is the
positive difference between these two approximations for the value of p?
Express your answer as a decimal to the nearest thousandths.[/quote]
ok, is this really hsb? come on. All you do is press them on your calculator, and if you can't use a calculator, then do it by hand!
If you can't do this by hand, then this shouldnt be in hsb.


Let the letters r and theta represent polar coordinates.  Write an equation using rectangular coordinates (x, y).

HINT IN THE BOOK

Use the following equations accordingly for each question as needed:

x = rcos(theta)

y = rsin(theta)

r^2 = x^2 + y^2

tan(theta) = y/x, where x does not equal zero.

(1) r = 4

(2) r = 3/(3 - cos(theta))

Solution

(1) $ r \equal{} 4$

$ \Rightarrow\sqrt{x^{2}\plus{}y^{2}}\equal{} 4$

$ \Rightarrow x^{2}\plus{}y^{2}\equal{} 16$

(2) $ r \equal{}\frac{3}{3\minus{}\cos{\theta}}$

$ \Rightarrow 3r\minus{}r\cos{\theta}\equal{} 3$

$ \Rightarrow 3\sqrt{x^{2}\plus{}y^{2}}\minus{}x \equal{} 3$

$ \Rightarrow 3\sqrt{x^{2}\plus{}y^{2}}\equal{} x\plus{}3$

$ \Rightarrow 9(x^{2}\plus{}y^{2}) \equal{} x^{2}\plus{}6x\plus{}9$

$ \Rightarrow 8x^{2}\plus{}9y^{2}\minus{}6x\minus{}9 \equal{} 0$


function
induction
algebra unsolved
algebra

find all function $f(x)$ satisfy $f: R\to R$and
$f(x-f(y))=4f(x) -f(y) -3x$$\forall x,y \in R$

f(x)=x

.............

when x_1 is not x_2,   f(x_1) is not f(x_2)( use the induction,)

and, f(x-f(x))=-3(x-f(x))

Let x-f(x)=A

so  f(A)=-3A

........

Let x=A+f(y)

so,f(y)=f(A+f(y))

Let y=A

f(A)=f(-4A)

so, A=0

so, f(0)=0

then we can find that f(x)=x

[quote="Kimjiman"]f(x)=x

.............

when x_1 is not x_2,   f(x_1) is not f(x_2)( it's easy to prove!)

and, f(x-f(x))=-3(x-f(x))

Let x-f(x)=A

so  f(A)=-3A

........

Let x=A+f(y)

so,f(y)=f(A+f(y))

Let y=A

f(A)=f(-4A)

so, A=0

so, f(0)=0

then we can find that f(x)=x[/quote]


$a\in\mathbb R$ is parameter and $x_1,$ $x_2,$ $x_3$ are roots of the equation:
$x^3-ax^2-(a+3)x-1=0.$
Prove that:
1. $x_1,$ $x_2$ and $x_3$ are different real numbers $\forall a.$
2. $\sqrt[3]{x_1}+\sqrt[3]{x_2}+\sqrt[3]{x_3}=\sqrt[3]{a+6-3\sqrt[3]{a^2+3a+9}}.$

EDIT:  I deleted my solution because it was wrong.  :oops:

[hide]suppose .  Then  and .  Combining these two equations gives , which has complex solutions.  Thus all 3 roots of the polynomial are complex, which is a contradiction since a cubic can have at most 2 complex roots.

suppose  and .  Then using viete's relations we have , , and .

After some algebra, eliminating  and , we have  which is a symmetric polynomial.  After dividing by  and making the substitution , we have , which has complex solutions in .  But this is a contradiction since , complex solutions must come in conjugate pairs.  So it's not possible that there's only 2 distinct roots.

So, all 3 roots must be distinct.  It remains to show that they're all real.  Suppose that there's 2 complex roots.  Then  for some positive real .  Equating coefficients gives , , and .  Solving for  gives .  But since , this implies  is negative which is a contradiction.  So it's not possible to have 2 complex roots.  Thus all 3 roots must be real.

for 2, i think this should pan out, using  on  and then on [/hide]

matt, you inexplicably decided that if two roots were equal they had to equal the (pre-selected) $a$, which is certainly not true.  cats..., you got through that part all right, but then you only considered the possibility of imaginary (not complex) roots.

Dear mathlinker, anyone can solve this problem?
I lost 1 hour to solve this problem, but i didn't success :(

What is the meaning of the first result?  Is it:

If $a_1 \neq a_2$, then $f(a_1,x) = f(a_2,y) = 0 \rightarrow x \neq y$?

Or is it:

$f(a, x) = f(a, y) = 0 \rightarrow x \neq y$  (i.e. never has double root for any value of $a$)?


logarithms
inequalities proposed
inequalities

Let $x$ be a real such that $0 < x < 1$. Manifestly $\sum_{n=1}^{N} \frac{x^n}{n(n+1)} < \frac 1x \sum_{n=1}^{\infty} \frac{x^{n+1}}{n(n+1)}$
But $(\sum_{n=1}^{\infty} \frac{x^{n+1}}{n(n+1)})'' = \sum_{n=1}^{\infty} x^{n-1} = \frac 1{1-x} \Longrightarrow \sum_{n=1}^{\infty} \frac{x^{n+1}}{n(n+1)}= x+(1-x)\ln (1-x)$
Put $x=\frac 12$ and here you are ..
 :cool:


geometry
geometric transformation
dilation
symmetry

[quote="zeus_three"]I have a question that has annoyed me for a while.

It has been said that the universe curves back on itself, such that travelling in one direction will eventually result in you returning to that very same position (albeit after a very long time).

The twin's paradox in special relativity is resolved by saying that the spacecraft has to slow down (i.e. decelerate), and then accelerate back towards earth - and hence, does not apply to inertial frames of references.

But, if the universe is curved, then there's no need to accelerate back to earth, i.e. after a (very) long time, the spacecraft will return to Earth anyway, and so the twin's paradox is not resolved.  What is the problem here?[/quote]In my opinion you are vesting too much significance in the twins paradox. The point of it is to highlight the phenomenon of time dilation, not to curse selected pairs of twins with some kind of magical aging spell that they can somehow then avoid by acceleration. The point was to select otherwise identical objects that by definition if Special Relativity did not apply would experience the same timeline and show what would occur to them because of time dilation.

First, because of the exigencies of the Lorentz transform no spacecraft will ever travel at relativistic speed. So becoming younger than your twin's children is something no one will ever have to worry about in a practical sense.

Second, the possibility of acceleration in itself does not disqualify a particular frame of reference as an inertial frame.  An inertial frame is one which travels along with its contents at a fixed velocity such that Newton's first law of motion is valid for said contents. If the spacecraft carrying the astronaut twin ever reaches a steady state velocity then it becomes an inertial frame, time dilation applies, the astronaut twin ages more slowly than his or her earthbound sibling, and the twins paradox is not resolved.

[quote="Dr. No"]If the spacecraft carrying the astronaut twin ever reaches a steady state velocity then it becomes an inertial frame, time dilation applies, the astronaut twin ages more slowly than his or her earthbound sibling, and the twins paradox is not resolved.[/quote]I would like to explain why is this a paradox and what does "resolved" mean. Firstly, the problem is [b]not[/b] that one of the twins will be older than the other one. As we all know, there's nothing wrong with it. The problem is that if we look at the whole situation from the reference frame of the Earth twin, he sees his sibling return home being younger than him. Then if we look at the problem from the reference frame of the astronaut twin, this astronaut twin returns home and finds his Earth twin being younger! Obviously, they can't be both younger than the other one, so where's the truth? The resolution which zeus_tree was talking about is that we say that the astronaut's reference frame is not inertial and therefore the way we used Lorentz transformations is incorrect (and it turns out that astronaut twin is the one who gets younger).

I'm interested in the curved universe theory though. Are there any reasons / proofs / arguments for this assumption?

[quote="bus"]The resolution which zeus_tree was talking about is that we say that the astronaut's reference frame is not inertial and therefore the way we used Lorentz transformations is incorrect (and it turns out that astronaut twin is the one who gets younger).[/quote]Agreed, it is the necessary acceleration of the astronaut twin that destroys the symmetry of the timelines of the two ... but I thought this should be obvious to everyone who has studied the problem. The astronaut twin is the one who makes the journey to Alpha Centauri, or wherever, so it is he who must accelerate to near luminal speed and thus it is he who ages according to the proper time.

So what happens if there's no acceleration? zeus_tree, you say that the astronaut twin could travel uniformly in one direction and after some time come back to Earth. I don't think there's any solution to the paradox in this case so the assumption that the universe is curved in this manner must be wrong, don't you think?

that's what i thought too, but i haven't seen anyone ever pose the question before.

I'm no physics expert but I have never heard anybody (seriously) talking about the curved universe, perhaps it's just something the philosophers made up? :)

Well, Hubble's law can be explained in terms of the curved universe.  And Einstein himself proposed the curved universe in his general relativity.

If the universe is curved, then isn't the traveling twin experiencing acceleration in order to come back to the same place in space?

If the traveling twin eventually reaches his original point in space due to the curvature of the universe, then isn't the twin technically accelerating?

My reply would be the twin is not in an inertial frame if in curved space - they are subject to gravitation and this is therefore outside the realm of special relativity...


suppose p is a prime such that (p-1)/4 and (p+1)/2 are also primes. show that p=13.

By changing $ \frac{p\minus{}1}4$ to $ p$ this is equivalent to asking for primes $ p$ such that $ p$,$ 2p\plus{}1$,$ 4p\plus{}1$ are all primes. If $ p\equiv 2\mod 3$ then the third will be a multiple of 3 and not 3. If $ p\equiv 1\mod 3$ (and $ p$ isn't 1, which isn't prime anyway) then the second number will be a multiple of 3 and not 3. So $ p\equiv 0\mod 3$ so $ p\equal{}3$ and $ 4p\plus{}1\equal{}13$.