Let  n  be a 5-digit number, and let  q  and  r  be the quotient and remainder, when  n  is divided by 100. For how many values of  n  is  q+r  divisible by 11?

[hide="Hint"]Try splitting it up - for example, how many numbers result when q is is 100 (its smallest possible value)?  How about 101?  Keep going and see if you notice a pattern.
[/hide]

[hide="Solution"]Using the method described in the hint, we will try some values of q:

If q = 100, r = 10, 21, ..., 98 => 9
If q = 101, r = 09, 20, ..., 97 => 9
..
q = 109, r = 10, ..., 89 => 9
q = 110, r = 00, ..., 99 => 10

So every 11 values of q, there are 10 values that each produce 9 solutions, and 1 that produces 10, for a total of 100 every 11.

Now, the values of q range from 100 to 999, so there are 900 numbers, i.e. 81 full sets of 11 numbers (81*11 = 891) plus 9 more numbers left over, all of which produce 9 possibilities.  So there will be 81*100 + 9*9 = 8181 total possibilities.
[/hide]

Another approach:
[hide]Try to show that q + r will be divisible by 11 if and only if n is divisible by 11.[/hide]

I think I see what you mean JBL.

[hide]$n=100q+r\Leftrightarrow\ n(mod11)\equiv(100q+r)(mod11)\Leftrightarrow\ n(mod11)\equiv100q(mod11)+r(mod11)$

But, $gcd(100,11)=1\Rightarrow\ 100q\equiv q(mod11)$. Therefore $n(mod11)\equiv (q+r)(mod11)$.

Now we count: the number of integers less than or equal to 10000 divisible by 11; the number of integers less than or equal to 99999 divisible by 11,and subtract them.

$\displaystyle \left \lfloor\frac{99999}{11} \displaystyle \right\rfloor -\displaystyle\left\lfloor\frac{10000}{11}\displaystyle\right\rfloor =9090-909=8181$.[/hide]

Right-o, good job :).  You do have a couple of errors, though:[hide]
[quote]$\displaystyle n=100q+r\Leftrightarrow\ n(mod11)\equiv(100q+r)(mod11)$[/quote] is not in fact true.  This statement is not an "if and only if" -- it only goes one way, but we only need it to go one way for the question.  Also, $100q\equiv q \mod 11$, but not because $\gcd(100,11) = 1$.  Instead, it's because $11 \mid 100 - 1$.  I imagine those were closer to typos than to deeper errors.

Also, you can make your tex look a little nicer -- use \gcd and \mod to unitalicize it.[/hide]

Oops, I see what you mean on the gcd. I wanted justification for cancelling off the 100; that $100\equiv1\mod 11$ was enough. :( Oy the shame! :blush:  . It was a pretty bad oversight on my part. For instance: $\gcd(5,3)=1$, but $5q\not\equiv q\mod 3$, rather $5q\equiv 2q\mod3$. $(q\neq0)$

The double arrow might be a worse blunder; since for instance $3\equiv (100\cdot 5+9)\equiv14\mod 11$, but $3\neq 100\cdot 5+9$.

JBL,thanks for the help,and sorry to all for any confusion.


[quote="pvthuan"]One co-operating author submitted the following inequality for me; he thought it was very hard one. But it is actually very easy. \[\sum_{\rm cyclic}x^{2}\sqrt{y^{2}-yz+z^{2}}\leq x^{3}+y^{3}+z^{3}.\] Do you find it just a cake? :lol:

He added the following one: \[\sum_{\rm cyclic}\frac1{\sqrt{x^{2}-xy+y^{2}}.\sqrt{y^{2}-yz+z^{2}}}\geq\frac{12}{(x+y+z)^{2}}.\] That co-operating begins with my inequality, then he gave a number of interesting results and you will find these results in my coming book. (He is one of students who sat for Vietnam IMO Selection Test 2006).[/quote]

The second is easy, because the same problem was in my book  :P  Notice that if $x\ge y\ge z$ then $x^{2}-xz+z^{2}\le x^{2}$, $y^{2}-yz+z^{2}\le y^{2}$, so we only need to check the case $z=0$. If that, the problem becomes very easy.

For the first, I handle it some seconds after, \[(\sum x^{2}\sqrt{y^{2}-yz+z^{2}})^{2}(\le \sum x^{3}) (\sum x(y^{2}-yz+z^{2}))\] and we are done, because \[\sum x(y^{2}-yz+z^{2})=\sum xy(x+y)-3xyz \le \sum x^{3}\] [quote="pvthuan"]That co-operating begins with my inequality, then he gave a number of interesting results and you will find these results in my coming book. (He is one of students who sat for Vietnam IMO Selection Test 2006).[/quote]
Dont say by that manner, please! Mathlinks is not as diendantoanhoc.net. Or I must think you like TP...   :(

Yes, that is right, hung. We need Cauchy and Schur only for the first.


What is the sum of the solutions for $ y$ for the equation $ 3(2^{2y\plus{}1})\minus{}(2^y)\minus{}1\equal{}0$?

[quote="erdogankerem123"]What is the sum of the solutions for $ y$ for the equation $ 3(2^{2y \plus{} 1}) \minus{} (2^y) \minus{} 1 \equal{} 0$?[/quote]

[hide]$ 6x^2\minus{}x\minus{}1\equal{}0$

$ (3x\plus{}1)(2x\minus{}1)\equal{}0$

$ x$ must be positive so $ x\equal{}\frac{1}{2}$

$ y\equal{}\minus{}1$[/hide]


Let $X_n$ be a sequence of real numbers such that $X_0,X_1 \geq 0$ and

$X_{n+2}= \frac{a + (b+X_{n+1}^p)^{1/p}}{c+(d+eX_{n}^q)^{1/q}}$ , where $a,b,c,d,e,p,q$ are positive and $c\geq 1$

Prove that $X_n$ is convergent . .. Could you generalize ? :)
 :cool:

are u sure that it isn't generalized ?  :P


find all real x, such that (13 + \sqrt x)^(1/3) + (13 - \sqrt x)^(1/3) is an integer

I think there is a mistake in the solution of Namdung .

We see : (13 + sqrt(15794))^1/3 + (13 - sqrt(15794))^1/3 = 0.34611...

Here is my solution:

We note that x>0.
Let y = (13 +  \sqrt x)^1/3 + (13 -  \sqrt x)^1/3
<=> y 3  = 26 + 3y(169-x)^1/3
<=> x = 169 - ((y 3 -26)/(3y)) 3
Hence : y in N  <=> y<5 ( It's easy to see that y \geq 5 => x<0 )
                      <=> y = 1,2,3,4 ( It's easy to see that y \geq 0)
So: x= 20188/27 , 196 , 123200/729 , 29645/216

Maybe I will have some terrible mistakes . Please read it carefully .
And I also cann't find the mistake in the solution of Namdung . Please show me it when you find out .


Hi everyone,

I am trying to get the nubmer in subsubsetion.
For that I am issue the command
\setcounter{secnumdepth}{-1}
or
\setcounter{secnumdepth}{3}

so when I insert

\subsubsetin{something}

to get

1.3.4 something.

but i don't get it.
i get only

something

Is there something I missing?
Thanks for any help
genick bar-meir
http://www.potto.org

It looks like you are misspelling subsubsection.  (Did LaTeX report an error?)  Change \subsubsetin to \subsubsection.

You are right that in post I miss spell the word
subsubsection.
however in the file I did spell it correct.
for example
\subsubsection{Gas Phase under Hydrostatic Pressure}

genick
http://www.potto.org

Are you setting secnumdepth in the preamble?  (Setting it to 3, not -1, is what you want.)  Can you give us a minimal, complete example that shows the problem?

:lol:
I found the extra command
which was the problem.
thanks,

I notice that you called Ravi B
Navier-Stokes Equations
you might be interested in my new book
on basic of fluid mechanics not yet done but
my other book on
compressible flow you can freely download from
http://www.potto.org

\setcounter{secnumdepth}{3}


abstract algebra
group theory
superior algebra
superior algebra solved

Thanks to fleeting guest, I found some very nice problems in some papers of Miklos Abert. Here is one of them:
   What is the maximal size of an abelian subgroup of $S_n$?

The maximal order is $3^{n/3}$.

Consider the orbits $O_{i}$ of $H$ (a maximal abelian subgroup). Then $H_{i}$ the image of $H$ in $\mbox{Sym}(O_{i})$, ... and an induction.


Find all functions $f$ defined on the integers and taking integer values, satisfying \[ f(a^3 + b^3 + c^3) = f(a)^3 + f(b)^3 + f(c)^3.\]

We used it for our correspondence program.
You may find a solution (in french, but it doesn't matter in that case) here :
http://www.animath.fr/tutorat.html#L'entra%EEnement%20de%20l'Olympiade%20fran%E7aise%20de%20math%E9matiques

See dossier 4 exercice 2.

Pierre.


Using the numbers 1, 3, 4 and 6 once each and the signs + , -, x and /, is it possible to make 24?

After about 10 seconds of getting lots of 12's and 48's, I gave up  :D and used a computer program to get  :oops:
[hide]
$ 24 \equal{} \frac{6}{1\minus{}3/4} \equal{} 6 \div (1\minus{}(3\div4))$
[/hide]

The 24 game seems to require more brute-forcing than most problems... I don't quite like that...

There is a 24 game marathon in this thread: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=209924[/url]


[quote="shinwoo"]Today is a very special day for those of you who belives in Einstein's Relativity Theory!

it's been exactly 100 years since the Relativity Theory was published..

so here's a little question

what's the speed of light?[/quote]

I don't know whether you are just kidding or you really want the answer
in all cases it's almost 300 000 km/s  :D  :D  :D

[quote="inteluser"]In a vacuum, of course.[/quote]
Of course, thank you inteluser :)

Yes; now I will stop being annoying and go to sleep, whence I shall dream of photons zipping through Cs gas.


Let $ a,b,c > 0$. Prove that

$ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \geq ( 1 \plus{} \sqrt {2} \plus{} \sqrt {3})\left( \frac {1}{a \plus{} \sqrt {2}b \plus{} \sqrt {3}c} \plus{} \frac {1}{c \plus{} \sqrt {2}a \plus{} \sqrt {3}b} \plus{} \frac {1}{b \plus{} \sqrt {2}c \plus{} \sqrt {3}a}\right)$

[quote="Nam Luu"]Can anyone give me the solution or the hint or sth ?:(  :(  :([/quote]

http://www.irgoc.org/viewtopic.php?f=10&t=12&sid=f2f035cada5360b5be65a6d0eb751071

It is kind of Cauchy- Schwartz application here.

\[ (a\plus{}\sqrt{2}b\plus{}\sqrt{3}c)\left(\frac{1}{a}\plus{}\frac{\sqrt{2}}{b}\plus{}\frac{\sqrt{3}}{c}\right) \ge (1\plus{}\sqrt{2} \plus{}\sqrt{3})^2\]

We are done now.


inequalities
LaTeX
inequalities proposed

[size=120][color=darkred]Let $a,b>0$ for which $(a-1)(b-1)>0$. Prove that $\boxed{\ a^b + b^a > 1 + ab + (1-a)(1-b)\cdot\min \{1,ab\}\ }$ .[/color][/size]

Latexed for you:

Reals $a,b >0$ are such that $(a-1)(b-1) > 0$.

Prove $a^b + b^a > 1 +a b + (1-a)(1-b)(\min\{1, ab\})$.

--
edit:  oops: i was wrong

You are wrong, Singular  :D . The correct LATEX transcription is : $a^b + b^a > 1 +a b + (1-a)(1-b)(\min\{1, ab\})$.
 :cool:

Your inequality is wrong. My inequality is correctly.

[quote="cezar lupu"]it is easy to prove that $a^b+b^a>=1+ab$ if $(a-1)(b-1)>0$[/quote]
how? I don't know how to deal with it when $a,b <1$.

[quote="cezar lupu"]Just use Bernoulli inequality.[/quote]
How?

[quote="cezar lupu"]Like this: $a^{1-b}=(1+a-1)^{1-b}<=1+(a-1)(1-b)=a+b-ab$ and so on...[/quote]
thx!

[size=120][color=darkred]Let $a,b>0$ for which $(a-1)(b-1)>0$. Prove that $\boxed{\ a^b + b^a > 1 + ab + (1-a)(1-b)\cdot\min \{1,ab\}\ }$ .

[u][b]Bernoulli[/b]'s inequality[/u] $:\ \left\{\begin{array}{cccc}
\blacktriangleright & 0<a<1 & \implies & \left( \forall \right) x>-1,\ (1+x)^a\le 1+ax\ .\\\\
\blacktriangleright & 1\ge a & \implies & \left( \forall \right) x>-1,\ (1+x)^a\ge 1+ax\ .\end{array}\right\|$[/color][/size]

[size=110][color=darkblue][b]A solution of the proposed problem.[/b]

[u]The first case[/u] $:\ 0<a,b\le 1\Longrightarrow \min \{ 1,ab\}=ab\implies$  $\left(\frac 1a\right)^b=\left[ 1+\left( \frac 1a -1\right)\right]^b\le \frac{a+b-ab}{a}\Longrightarrow a^b\ge \frac{a}{a+b-ab}\ .$

Analogously, $b^a\ge \frac{b}{a+b-ab}$ . Thus,  $a^b+b^a\ge \frac{a+b}{a+b-ab}=1+ab+\frac{ab(1-a)(1-b)}{a+b-ab}.$ Since $a+b-ab=a+b(1-a)>0\ ,$

and$\frac{ab}{a+b-ab}\ge ab\Longleftrightarrow (1-a)(1-b)\ge0$ obtain that $\frac{ab(1-a)(1-b)}{a+b-ab}\ge ab(1-a)(1-b).$

Therefore, $a^b+b^a\ge 1+ab+ab(1-a)(1-b)=1+ab+(1-a)(1-b)\cdot \min \{1,ab\}.$

[u]The second case:[/u] $1\le a,\ 1\le b \Longrightarrow \min \{1,ab\}=1.$ Thus, $a^b\ge 1+b(a-1)$, i.e. $a^b\ge 1+ab-b$.

Analogously, $b^a\ge 1+ab-a$. Thus, $a^b+b^a\ge 1+ab+(1-a)(1-b)=1+ab+(1-a)(1-b)\cdot \min \{1,ab\}.$[/color][/size]

How do you prove that $a^b+b^a>1+ab$ for any $a,b>0$?

In general, if $0<a<1<b$ it is not true. For example:
$a=\frac 12,\ b=2\Longrightarrow a^b+b^a=0.25+\sqrt 2\ \overline {\underline {\left| \ <\ \right| }}\ 2=1+ab$.

Thx. I don't know why I thought it is true :)

Yes, Levi you are right. The right inequalityis:

  If $x, y$ are both $\in (0,1)$ OR $x, y>1$ then the following inequality holds: \[ x^{y}+y^{x} >1+xy. \]


geometry
ratio
Pythagorean Theorem
geometry proposed

Hi.

I show you the Archimedes' Book of Lemmas.

15 problems to lift up our geometry skills.

Once solved the proposition, I'll post the next one.

[u][b]Proposition 1[/b][/u]

[img]http://img442.imageshack.us/img442/7979/archboolem01hg1.gif[/img]

Cya!!

[hide="Hint"]
Prove that both centers of circles and tangency point $A$ are collinear[/hide]

this is a very well known problem...i think you can just assume that tangent circles have collinear centers...if not, just note that both lines connecting the centers to the point of tangency are perpendicular to the tangent line at the same point, so they are the same line

basically you can then extend that line connecting centers, then you get two similar right triangles, so AC/AE=r/R where those are the radii of the circles, the parallel lines follow from SAS similarity

This result is generally described as: two tangent circles are homothetic with respect to the point of tangency...basically meaning that the circles are dialations of each other with respect to the point of tangency

Cool!!!

Keep going...

[u][b]Proposition 2[/b][/u]

[img]http://img101.imageshack.us/img101/8739/archboolem02zm3.gif[/img]

Cya!!

Let $AC$ intersect the circle again at $G$. ($G$ between $F$ and $C$),
Extend $ED$ to hit the circle again at $H$.

$\frac{DG}{\boxed{DF}}=\frac{AH}{AF}=\frac{AD}{AF}$ since $\triangle AHF\sim\triangle DGF$, and $AH=AD$

$\frac{AF}{AB}=\frac{\boxed{FE}}{BG}$ since $\triangle AFE\sim\triangle ABG$.

hence, $\frac{DG}{\boxed{DF}}*\frac{\boxed{FE}}{BG}=\frac{AD}{AF}*\frac{AF}{AB}$

--
Lemma: $AD*GB=DG*AB$

$\frac{AD}{DG}=\frac{AC}{DC}$ and $\frac{AB}{BG}=\frac{CB}{AC}$ since $\triangle ADC\sim\triangle DGC$ and $\triangle ABC\sim\triangle BGC$, since $CD=CB$, that finishes the lemma
--

it follows that $DF=FE$!

Right.

Let's get it on!!

[u][b]Proposition 3[/b][/u]

Fixing image...

 :maybe:

denote $O$ as the center, then $\angle AOC=\angle ABF$ (using inscribed angle theorems

also note that $\angle AFB=90=\angle CDO$, so $\triangle AFB\sim\triangle CDO$, so

$\frac{AF}{CD}=\frac{FB}{DO}=\frac{AB}{CO}=2$ (diameter/radius)

now, $AB=2(AO)=2AD-2DO=AD+DE+FB$

so $AB-AE=FB$,

$BE=FB$!!

Yeah!!!

It's good to post this beatiful problems.

Now another clasic.

[u][b]Proposition 4[/b][/u]

[img]http://img211.imageshack.us/img211/263/archboolem04vl6.gif[/img]

Bye!!

P.D.: if some people don't know what does mean "arbelos", it means [u]shoemaker's knife[/u] in Greek.

denote the radius of the largest circle $a$, and the radii of the two tangent circles with centers on AB, $b$, $c$

(note $a=b+c$)

arbelos= $\frac{\pi}{2}(a^{2}-b^{2}-c^{2})=\pi(bc)$

since $\triangle ADC\sim\triangle CDB$, $CD^{2}=AD*DB$

$\left(\frac{CD}{2}\right)^{2}\pi=\pi*\frac{AD}{2}*\frac{DB}{2}$

area of the yellow circle $=\pi*b*c$

as desired

[yes these are very nice problems! i am amazed that math from thousands of years ago can be so interesting]

Got it!!

Here we've got another one.

[u][b]Proposition 5[/b][/u]

[img]http://img297.imageshack.us/img297/5205/archboolem05xm1.gif[/img]

just wondering; where are you getting the pictures from? they're very nice; as are the problems :D

[quote="pkerichang"]uhh are we suppose to hide??? I'll just do it anyways:

[hide]
draw line $EG$ and $FH$ such that $G$ and $H$ are on $AB$ and $EG$ and $FH$ are both perpendicular to $AB$.  Also call the center of semicircle $AB$ $O_{1}$, center of $AD$ $O_{2}$ and center of $DB$ $O_{3}$.  Pythagorean on $EGO_{1}$ and $EGO_{2}$ yields $EG^{2}= (R-r)^{2}-(R-r-2b)^{2}= (r+a)^{2}-(r-a)^{2}$, which yields $b(R-b) = (a+b)r \implies ab = (a+b)r$.  ($r$ is the radius of circle $E$.).  Similarly, if $r'$ is the radius of circle $F$, then $(a+b)r' = ab$.  So they're equal.
[/hide]
Let me guess, the next problem is going to be that the external tangent of semicircle $AD$ and semicirlce $BD$ bisects $CD$ :D[/quote]

lol i was going to distance formula brute force too...i wonder if there is a better way...I was thinking of inversion, but that does not seen to help that much...

[quote]just wondering; where are you getting the pictures from? they're very nice; as are the problems[/quote]
Made'em by Corel Draw  :ninja:

[quote="pkerichang"]
draw line $EG$ and $FH$ such that $G$ and $H$ are on $AB$ and $EG$ and $FH$ are both perpendicular to $AB$.  Also call the center of semicircle $AB$ $O_{1}$, center of $AD$ $O_{2}$ and center of $DB$ $O_{3}$.  Pythagorean on $EGO_{1}$ and $EGO_{2}$ yields $EG^{2}= (R-r)^{2}-(R-r-2b)^{2}= (r+a)^{2}-(r-a)^{2}$, which yields $b(R-b) = (a+b)r \implies ab = (a+b)r$.  ($r$ is the radius of circle $E$.).  Similarly, if $r'$ is the radius of circle $F$, then $(a+b)r' = ab$.  So they're equal.[/quote]
Good.

[quote="pkerichang"]uhh are we suppose to hide??? I'll just do it anyways[/quote]
Is not necessary.

[quote]Let me guess, the next problem is going to be that the external tangent of semicircle $AD$ and semicirlce $BD$ bisects $CD$ :D[/quote]
Sorry, you dind't guess  :ninja:

[u][b]Proposition 6[/b][/u]

[img]http://img513.imageshack.us/img513/5163/archboolem06ix4.gif[/img]

This is a very interestin' problem, jeje.

I was checking the answer for the proposition 3, and I found this.

[quote="Altheman"]$AB=2(AO)=2AD-2DO=AD+DE+FB$[/quote]
Must be $2\overline{AD}+2\overline{OD}$

Bye.

Cool, but I'd like a proof without Descartes' Circle Theorem.

Once solved for two ways, I'll post the next problem.

ok...i have a partial solution...assume that DE is parallel to AB,

let C (AB) be tangent to C(DE) at S
C(AC) "" C(DE) at T
C(BC) "" C(DE) at R

we know that SDA, SEB, STA, DTC, XRB, ERC are collinear since tangent circles are homothetic with respect to the point of tangency... (i.e proposition 1)

WLOG CB=1, AC=r, DE=x,

since AE is perpendicular to DC, (r+x)^2=DB^2+EC^2
likewise, (1+x)^2=DB^2+EC^2

since EC is perpendicular to AE, $r^{2}+x^{2}=AD^{2}+CE^{2}$, likewise, $x^{2}+1^{2}=DC^{2}+EB^{2}$

then we get $2x+2rx-DB^{2}-AE^{2}=AD^{2}+EB^{2}$

now i can handle AD, and BE because of the similarity...but AE, DB, are harder to deal with...I will work on this later...i have a lot of school work to do... :mad:

[quote="Geómetra"][quote="pkerichang"]$EG^{2}= (R-r)^{2}-(R-r-2b)^{2}= (r+a)^{2}-(r-a)^{2}$, which yields $b(R-b) = (a+b)r \implies ab = (a+b)r$.[/quote]
I'd like you explain what represents $a$ and $b$.[/quote]

I believe $a$ is AC, and $b$ is CB.

now...it's BRUTE FORCE TIME!!!

let $O_{1}$ be center of semicircle AC, $O_{2}$ be center of semicircle CB, and $O_{3}$ be center of circle with diameter DE.  Look at triangle $O_{1}O_{2}O_{3}$, and WOLOG $CB = 1$, so that $AC = r$.  let the radius of circle with center $O_{3}$ be $x$, and yes, you guessed it, distance brute force...

If $O$ is the center of the semicircle AB, then consider the cevian $O_{3}O$, by some inspection we have $OC = 1$, and $OO_{2}= r$, so by Stewart's we have $(r+1)((r+1-x)^{2}+r) = (x+r)^{2}r+(x+1)^{2}$.  Now through painful yet doable algebra (I have Mathematica so I don't care :rotfl: ) we get $x = \frac{r^{2}+r}{r^{2}+r+1}$, divide by $r+1$, the radius of semicircle AB, to get the result.

I see..

But we don't get the idea...

Try to solve it without [i]powerful theorems[/i], cuz is much beauty with basic geometry.

Once done, I'll post the next one.

[u][b]Proposition 7[/b][/u]

[img]http://img171.imageshack.us/img171/459/archboolem07xs5.gif[/img]

draw the radius of the big circle from a vertex of the square, then draw the radius of the small circle to the tangent point.  Then we see that we have a 45-45-90 triangle, and therefore the radius of the big circle is $\sqrt{2}$ times the radius of the small circle.  Therefore, the area ratio is $\sqrt{2}^{2}= 2$.

No one can solve the other one w/o distance brute force??? I'm quite interested... :maybe:

[quote="pkerichang"]...No one can solve the other one w/o distance brute force??? I'm quite interested... :maybe:[/quote]

OK. Denote $a = AC,\ b = BC,\ d = DE$ diameters of the circles (P), (Q), (R), respectively. Let a normal to AB at C meet the big circle (O) with diameter $AB = a+b$ at F, F'. Then $c = CF = \sqrt{ab}.$ Inversion in a circle (C) with center C, radius c = CF, and negative power $-c^{2}$ carries the big circle (O) into itself, because it intersects the inversion circle (C) in diametrically opposite points F, F'. The circles (P), (Q) passing through the inversion center C go into parallel tangents $t_{B}\parallel t_{A}$ of (O), both perpendicular to AB. The circle (R) goes into a circle (R') tangent to $(O),\ t_{A},\ t_{B},$ i.e., congruent to (O), with diameter d' = a + b. The inversion center C is a similarity center of (R), (R'), with similarity coefficient equal to the power of inversion divided by the power of the inversion center C to the inverted circle (R'):

$\frac{DE}{AB}= \frac{d}{a+b}= \frac{d}{d'}= \frac{c^{2}}{p(C, (R'))}= \frac{c^{2}}{CR'^{2}-\left(\frac{d'}{2}\right)^{2}}= \frac{c^{2}}{CO^{2}+OR'^{2}-\left(\frac{d'}{2}\right)^{2}}=$

(Pythagorean theorem for the right $\triangle COR'$ was used.)

$= \frac{ab}{\left(\frac{b-a}{2}\right)^{2}+(a+b)^{2}-\left(\frac{a+b}{2}\right)^{2}}= \frac{4ab}{(b-a)^{2}+3(a+b)^{2}}= \frac{ab}{a^{2}+ab+b^{2}}=$

$= \frac{\frac{a}{b}}{\left(\frac{a}{b}\right)^{2}+\frac{a}{b}+1}= \frac{r}{r^{2}+r+1}$

[u][b]Proposition 8[/b][/u]

[img]http://img412.imageshack.us/img412/4615/archboolem08ig6.gif[/img]

If $\overline{CD}$ be any chord of a circle whose center is $O$, and if $\overline{CD}$ be produced to $A$ so that $\overline{AD}$ is equal to the radius; if further $\overline{AO}$ meets the circle in $B$, then the $\measuredangle \ BOC$ will be equal to three times the $\measuredangle \ BAC$

I wonder why Geómetra isn't posting?
[hide="here is a written out proof to explain yettis diagram just in case"]
[size=125]Since $|DO|=r=|AD|$ by the Base Angles Thm., $m(\angle A)=\alpha=m(\angle AOD)$. Since $m(\angle CDO)$ and $m(\angle A+\angle AOD)$ and supplementary to the same angle, they are equal i.e. $2\alpha = m(\angle ODC)$. Since $|DO|=r=|CO|$, by the Base Angles Thm., $m(\angle OCD)=m(\angle ODC)=2\alpha$. Since $m(\angle AOD+\angle rOC)$ and $m(\angle OCD+\angle ODC)$ are supplementary to the same angle, they are equal hence $2(2a)=a+\angle rOC\implies m(\angle rOC=3a)$, which is basically as close as one gets to a proof without algebra.[/size]
[/hide]

[color=darkblue][b]Good show, Geometra ![/b]

The your problems are very nicely and interestingly.

[u]There is a book with like problem ? [b]Thank you a lot ![/b][/u][/color]


quadratics conics parabola function inequalities algebra

Find all values of $k$ such that the roots of the quadratic equation $4x^2-8kx+k=0$ has two distinct real solutions between $0$ and $1$.

$r1+r2\leq 2$ and $r1 \cdot r2\leq \frac{k}{4}$, but in that case, $k\leq 1$, bt I really don't know where to go from there....

[hide]These conditions must be satisfied simultaneously:

- the vertex of the parabola must be between $0$ and $1$ - this yields $0<k<1$
- the values of the function must be positive in $0$ and $1$ - $(k>0)\wedge (k<\frac{4}{7})$
- the discriminant must be positive - $64k^2-16k>0\ \Rightarrow\ (k<0)\vee \left (k>\frac{1}{4}\right )$

Finally $k\in\left (\frac{1}{4},\frac{4}{7}\right )$.[/hide]

How do you hide solutions?

$4x^2-8kx+k = 0$
$x = \frac{8k\pm\sqrt{64k^2-16k}}{8} = k\pm\frac{\sqrt{4k^2-k}}{2}$
Since the roots are distinct, $k \not=0,\frac{1}{4}$
$k+\frac{\sqrt{4k^2-k}}{2} \le 1$ and $k-\frac{\sqrt{4k^2-k}}{2}\ge0$
$\frac{4k^2-k}{4} \le 1+k^2-2k$ and $k^2 \ge \frac{4k^2-k}{4}$
Therefore, $k\ge0$ and $k \le \frac{4}{7}$
The roots must also be real so $4k^2k-k \ge 0$, therefore $k \ge \frac{1}{4}$ or $k \le 0$
Combining all of these inequalities together,
$k\in(1/4,4/7]$
I guess if between means exclusive, then aappaa is more correct.

Press the "Hide" button right above the text box. You can also write

[code]
[hide]
Whatever you want to hide goes here.
[/hide]
[/code]

Yeah, $\left(\frac{1}{4},\frac{4}{7}\right)$ is correct. Nice solutions, aappaa and FMako ;)

Sorry that the question was a bit ambiguous.


Well, I decided I'll have like no rules. (that is to a certain person)
 Anyone interested in this test tell me.

The test would consist of 4 rounds of 4 questions(in June of course)
this test is recommended for but no limited to people who are tired of losing
This test will be so easy, that I will have to figure out how I will not have so many people all winning it(assuming I get takers.)

Notice that I would give people a week to do the test and extend if needed and I would most definitely try to stay on top of it(hence the only 4 week test period.)

So...anyone interested post me so that I can see if I will actually get this to go....

Anyone interested in writing really easy questions meaning below state mathcounts level tell me.

I'm in!

The message is too small. Please make the message longer before submitting.

[quote="CircleSquared"]I'm in too!

I'll write the uber easy questions....
I think....[/quote]
that's good...i will see when i am done with my last finals how much help i might need

Don't apologize, this contest is for people who are tired of losing  :D

even though I'm not really tired of losing  :rotfl:

i'm not sure if when people said "i lost" they mean count me in or if they find this humorous or they just have no idea what they are doing

Count me in!

I think they were referring to The Game.  If you don't know what that is, [hide="click here"]The Game is a game that you never stop playing.  Whenever you think of The Game, you lose The Game and therefore say "I lost!"[/hide]  If this is what you were referring to, I lost also.

[quote="Temperal"]I don't know whether I lost. Why is The Game capitalized? Is it sacred?

Count me in.[/quote]

The Game is the one against Harvard and Yale.
I am guessing these people have a different one.

I'm in this game that helps in absolutely no way.

Uber is pseudo-German (I think) for " a lot"

[quote="alanchou"]Congratulations, HoD, you found the Caps Lock key (I think).

When's this going to start?[/quote]

VERY SOON. BY TOMORROW OR THE NEXT DAY DEPENDING ON WHETHER I FIND IT APPROPRIATE TO SELECT *CERTAIN* QUESTINOS

I FOUND THAT KEY TOO!!!

okay since this is for losers the first round should be ummm easy
anyone who gets a perfect is eliminated
so don't search the internet for answers(if you are that dumb :P ) and maybe umm for the sake of keeping it going get some wrong... :wink:
MAN I MESSED UP ON MY FONTISH THINGS WHEN I CONVERTED IT!!!1!
Never mind...it's still good, eh?
DEADLINE: June 8th(see? enough time to brute force it)
PM for any extensions
you can PM me with answers even if you didn't say you'd do it
eh i can't seem to attach .pdf so .jpg it is

oh, edit you may use calculators that aren't made by ti that's weird, eh? :rotfl: i can't quite make sure that you did or did not....

[quote="xpmath"]So we don't want to get a perfect? Too bad...[/quote]
uh i never said the point was to be good at it...otherwise i would've posted elsewhere  :P

umm if no one pms me answers by at least the tenth other then xpmath we will have the ultimate winner/loser already

[quote="Klebian"]doesn't 3 have more than 2 values?[/quote]

[hide="Ph33r m3"]well if it was two digit numbers we'd have four: 23,37,53,73
but we have to add 100s

so 123=3*41
137 is prime
153=3^2*17
&, finally
173 is prime

add the two primes up but I've said too much[/hide]

so umm no one click this unless you want to either fail or not participate....

read the problem, it doesn't specify that the numbers themselves have to be prime

:o ok then

I will now accept multiple answers.

I forgot all about this. I'll play next round.

NO!!! YOU HAVE TO PLAY THIS ONE OTHERWISE WE WILL ALREADY BE DONE!!!! PLEASE!!!!

Answers That I was supposed to PM you:
[hide]1. 43
2. 75
3. 34
4. 3209[/hide]
 :first:  :first:  :first:

Fine, I'll just PM you a bunch of answers that I know are wrong, because I'm too lazy to actually do them.

ok i had 4 submissions.

2 didn't notice that they weren't pming me or they did it on purpose.

xpmath got correct except for instead of units digit showed units and tens so i'll count that as wrong just for fun(but really that would be like ex cr to some people)

everyone else FAILED(and temperal answered 5 questions!!!) YAYAY YIPPEEEE!!! w00t

oh no i must go and write a new set

oh no i will not be able to have 4 of them at this rate

list
(that is so sucky)

temperal  5 incorrect

asmallgoat
cincodemayo5590 4 incorrect

xpmath    3 correct

I did that on purpose. Just so you know, I will alternate between getting 3 right and 4 wrong.


Let $\{x\}$ denote the closest integer to $x$ (using the standard rounding conventions). Define $f(n) : = n+\{ \sqrt{n}\}$. Prove that, for every positive integer $m$, the sequence
\[f(m), f(f(m)), f(f(f(m))) \]
never contains the square of an integer.

[hide] it is sufficient to show that f(k) can't be a square for any integer k.

n^2 <= k <= n^2 + 2n + 1

n^2 + n <= f(k) <= n^2 + 3n + 2
the only perfect square in that range is  n^2 +2n +1 = (n+1)^2

now if k >= n + 1/2 we round up and if its less we round down, however:
(n+1/2)^2 =  n^2 +n +1/4
so f(n^2 + n) = n^2 + 2n
and f(n^2+ n +1) = n^2 + 2n +2

so the perfect square is skipped, and thus f(k) can never be a perfect square[/hide]


ARML
geometry
analytic geometry
3D geometry
algebra
polynomial
abstract algebra

dude yeah did I already tell you guys I am hopefully COMING BACK TO ARML OH MAN I AM PUMPED

also what apparently WOOT grading has still not started blargh

anyway yeah I couldn't really think of a better place to put a "HEY GUYS WHAT'S GOING ON" kind of message

but since it's in this thread, maybe I should give y'all a problem?

ok ok

I've thought of a fun one

given some field $ k$, define $ \mathbb{A}^{n}$ as $ n$-dimensional affine space, i.e. isomorphic to $ k^{n}$ (it's basically the same thing but we give it a different name. Then define $ n$-dimensional projective space as $ \mathbb{P}^{n}\equal{} (\mathbb{A}^{n\plus{}1}\setminus\{0\})/{x ~\lambda x}$, i.e. the quotient of $ n\plus{}1$-dimensional affine space (minus $ 0$), with the equivalence relation $ [x_{0}, x_{1},\cdots x_{n}] \equal{} [cx_{0}, cx_{1},\cdots, cx_{n}]$ for some constant $ c\in k^{*}$. (All of this is probably on Wikipedia if you don't like my explanations.)

Ok!

So now, consider $ \mathbb{P}^{4}$. Define a subset $ X$ as the set of points $ [p_{0},\cdots, p_{4}]$ such that $ \sum p_{i}\equal{}\sum p_{i}^{3}\equal{} 0$. Find how many straight lines lie in $ X$ (i.e. things isomorphic to $ \mathbb{P}^{1}$).

Also ok yea this is PROBABLY NOT an ARML problem (it was actually on one of my last homeworks). But you guys haven't been doing anything here for a couple weeks anyway. So ha.

[quote="MysticTerminator"]Then define $ n$-dimensional projective space as $ \mathbb{P}^{n} \equal{} (\mathbb{A}^{n \plus{} 1}\setminus\{0\})/{x ~\lambda x}$, i.e. the quotient of $ n \plus{} 1$-dimensional affine space (minus $ 0$)

...

So now, consider $ \mathbb{P}^{4}$. Define a subset $ X$ as the set of points $ [p_{0},\cdots, p_{4}]$ such that $ \sum p_{i} \equal{} \sum p_{i}^{3} \equal{} 0$. Find how many straight lines lie in $ X$ (i.e. things isomorphic to $ \mathbb{P}^{1}$).
[/quote]

wait, I am confused... is $ \mathbb{P}^1$ like... a plane? huh? (sorry, I'm probably being, in jb05's words, "dense")

$ \mathbb{P}^1$ is a "line" in $ k$-space together with a point at infinity.  If $ k \equal{} \mathbb{C}$, $ \mathbb{P}^1$ is a sphere.  If $ k \equal{} \mathbb{R}$, $ \mathbb{P}^1$ is a circle.  I don't think the geometry has much to do with the problem.  Anyways nobody on this forum knows any algebraic geometry.  Maybe I'll look at this though... I haven't yet.

What Arnav wants is the number of subsets of $ \mathbb{P}^4$ (as he's defined it in terms of equivalence classes of coordinates) that
a) Have only elements satisfying both of his conditions on sums of powers of coordinates
b) Map bijectively onto $ \mathbb{P}^1$ via a map that preserves subtraction

The way I was thinking of it before I gave up was: given three non-colinear points in $ \mathbb{R}^5$ with sum of coordinates and sum of cubes of coordinates 0, we want all linear transformations of them to have sum of coordinates and sum of cubes of coordinates 0. In this case, we will have something isomorphic to $ \mathbb{P}^1$. Or: given $ x_0$ through $ x_4$, $ y_0$ through $ y_4$, and $ z_0$ through $ z_4$, we want $ \sum P(x_i,y_i,z_i)$ for each odd polynomial P of degree 3 or less. Maybe this doesn't help, but I am pretty sure it is equivalent to the original problem.

While we're dealing with advanced problems, does anyone know how to determine whether the additive groups of reals and complexes are isomorphic? I've done a good bit of research on infinite Abelian groups just for this problem, but I still don't see anything useful. I don't even know any good defining properties of the reals without invoking an ordering.

Answer to Berman's question:

[hide]
Choose a basis for $ \mathbb{C}$ considered as a $ \mathbb{Q}$-vector space.  Similarly choose a basis for $ \mathbb{R}$.  The two bases have the same cardinality, so there is a bijection between their elements.  We can extend this bijection to get a $ \mathbb{Q}$-linear isomorphism of $ \mathbb{C}$ onto $ \mathbb{R}$.  This is a vector space isomorphism, and so in particular is an isomorphism of additive groups.[/hide]

[quote="n^4+4"]
Choose a basis for $ \mathbb{C}$ considered as a $ \mathbb{Q}$-vector space.  Similarly choose a basis for $ \mathbb{R}$.  The two bases have the same cardinality, so there is a bijection between their elements.  We can extend this bijection to get a $ \mathbb{Q}$-linear isomorphism of $ \mathbb{C}$ onto $ \mathbb{R}$.  This is a vector space isomorphism, and so in particular is an isomorphism of additive groups.[/quote]

1. I don't see why the arbitrary bijection between the bases necessarily is well-defined if we just throw in an operation. Reals have multiple representations as linear combinations of elements of the basis since the basis is infinite.

2. You claim that there is only one divisible, torsion-free group of each cardinality greater than $ \aleph_0$?

dude what? if they had multiple representations as linear combinations of the basis, it would not be a basis

a basis is kind of defined as OH MY GOD UNIQUE LINEAR COMBINATIONS WITH RESPECT TO THE BASIS

and it is trivial that any map from a basis induces a map on the whole space

you seem to not actually believe the Hamel basis is a basis

in spite of it being defined as a basis

also guys? my question is not actually algebraic geometry? otherwise I would not have given it to you?

dude yeah I learned what projective spaces were at MOP

well no I learned them by myself

but my point is they are taught at MOP?

[quote="MysticTerminator"]also guys? my question is not actually algebraic geometry? otherwise I would not have given it to you?[/quote]

That is why I formulated it as basic algebra. I still don't see how to solve it though. And yeah I didn't really believe it was a basis. In fact, I don't really believe R has a basis wrt Q. Which means... I need to learn more math.

look up zorn's lemma and prove every vector space has a basis

Hmm yeah that is pretty obvious using zorn isn't it... Oh well.

ok now prove zorn's lemma is equivalent to AC

if you do this post it here because I want to see

Well I've seen AC->Zorn before. Here's that and my own Zorn->AC.

[hide="Axiom of Choice implies Zorn"]Suppose all chains have upper bounds. Since, for the sake of contradiction, these elements are not maximal for the set, there are larger elements. Consequently, each chain has an upper bound not in itself. Choose an upper bound $ b_S$ for each well-ordered set $ S$ with $ b_S\notin S$. For a set $ C$ and element $ c$, let $ C(c)$ be the set of all things less than $ c$ in $ C$. Consider the set $ C^*$ of all $ C$ containing $ b_\emptyset$ such that $ b_{C(c)} \equal{} c$ for all $ c\in C$. Certainly, $ \{b_\emptyset\}$ is such a $ C$, so $ C^*$ is nonempty. It is pretty easy to prove that if $ C,D\in C^*$, then WLOG $ C \equal{} D(d)$ for some $ d\in D$ (basically because $ C$ and $ D$ are well-ordered with minimum $ b_\emptyset$ and then some sort of induction). This makes $ X \equal{} \bigcup C^*$ well-ordered and in fact an element of $ C^*$. However, $ X\cup\{ b_X\}\notin C^*$ is a contradiction, thus proving Zorn's Lemma for well-ordered chains. Furthermore, it is obvious that every chain has a well-ordered subset with the same upper bounds, which finishes it.[/hide]
[hide="Zorn implies Axiom of Choice"]Consider all subsets of $ X\neq\emptyset$ which satisfy AC. Define the partial ordering $ \subseteq$ on these subsets in the usual way. There is at least one subset (containing a single element) which satisfies AC. Given a chain $ \{ C_i\}$, consider $ \bigcup C_i$. For every subset $ S\subseteq \bigcup C_i$, there is some $ i$ such that $ S\cap C_i\neq\emptyset$. Then we may choose an element of $ S\cap C_i$ for each $ S$ since $ C_i$ satisfies AC. Consequently, $ \bigcup C_i$ satisfies AC, so every chain has an upper bound. By Zorn, there is a maximal subset $ M\subseteq X$ satisfying AC. If $ M \equal{} X$, we are done. Otherwise, there is some $ x\in X$, $ x\notin M$. Then $ M\cup\{x\}$ certainly satisfies AC, contradicting the maximality of $ M$. So $ X \equal{} M$ and all sets $ X$ satisfy the Axiom of Choice.[/hide]

Here's a problem: prove that there is a subset of the plane intersecting each line in exactly two points.


function
inequalities
integration
trigonometry
calculus
calculus computations

Prove the inequality: \[\int_{0}^{\pi \over 2}\sqrt{\sin x}\left(2+3\sqrt{\sin x}\right)\left(7-4\sqrt{\sin x}\right)dx <{27\pi \over 5}.\]

The reverse inequuality holds!  :wink: \[\int_{0}^{\pi \over 2}\sqrt{\sin x}\left(2+3\sqrt{\sin x}\right)\left(7-4\sqrt{\sin x}\right)dx >{27\pi \over 5}.\] [hide]The function $y\mapsto y(2+3y)(7-4y)$ is increasing on the interval $(0,1)$.
[/hide]
[hide]
For $x\in (0, \pi/2)$, \[\sin x \geq \frac{2}{\pi}x.\] Therefore, \[\sqrt{\sin x}\geq \sqrt{\frac{2}{\pi}x}.\] [/hide]
[hide]
Conclude that the left hand side is at least \[\int_{0}^{\frac\pi2}\sqrt{\frac{2}{\pi}x}\left(2+3 \sqrt{\frac{2}{\pi}x}\right)\left (7-4 \sqrt{\frac{2}{\pi}x}\right) dx.\] Compute the integral above.
[/hide]


geometry
rectangle
combinatorics unsolved
combinatorics

S is a finite set of closed intervals, whose union covers [0, 1] completely. What's the biggest k (positive real number) such that we can always find D, a subset of disjoint elements in S, such that the length of D is at least k.

I think $k={1 \over 2}$ is the best possible.  The example $[0,{1 \over 2}] \cup [{1 \over 2}, 1]$ shows you can't do any better.

To show ${1 \over 2}$ is always achievable, assume the contrary.  Then there must be a minimal counterexample, that is to say a counterexample involving the fewest number of intervals.  If any interval in this counterexample were to be contained in the union of the others, we could drop it to get a smaller counterexample.  Therefore we can assume that in our minimal configuration each interval $I$ contains a point $x_I$ which is unique to it.

Order the sets $A_1, A_2, \dots , A_n$ such that $x_1<x_2< \dots , x_n$.  The only sets which can possibly intersect $A_j$ are $A_{j-1}$ and $A_{j+1}$, as the points $x_{j-1}$ and $x_{j+1}$ separate $A_j$ from the remaining intervals.  It follows that the sets $A_1 \cup A_3 \cup \dots$ and $A_2 \cup A_4 \cup \dots$ are both the union of disjoint elements of $S$.  These two sets cover the interval, so one of them has total length at least  ${1 \over 2}$

you are right kevinatcausa, the key is to rank the intervals in some order, and your x_i did exactly that! Great job. :)

p.s. Just for the rigor of the solution, you might as well define x_i to be the smallest number in the interval A_i that is not contained by any other element in S.

The catch is, such a smallest number might not exist (for example if the intervals are $[0,.4], [.3, .7], [.6, 1]$, then $x_2$ has to be chosen from $(.4, .6)$.  There's only a finite number of intervals, so it doesn't really do any harm to just pick a point  arbitrarily from each.

Another question: What happens in higher dimensions, i.e. for covering $[0,1]^d$ with closed $d$-dimensional rectangles?  It's pretty easy to get an upper bound of $1/2^d$ in this case, but I don't know if it's always achievable


2008 School Sprint Round #29:

I took this test today during Mathcounts practice and the answer key says that the answer is sqrt 22 whereas I got the answer as sqrt 21 using the Pythagorean theorem.

I don't know how to get the image but here is the problem:
[b]In the figure shown, AC = 13 and DC = 2 units. What is the length of segment BD? Express your answer in simplest radical form.[/b]

http://216.235.203.102/NETCOMMUNITY/Document.Doc?id=177

P.S It is a right triangle.

Thanks,
Canton

[hide]We have $ \triangle{ABD}\sim\triangle{ACB}\sim\triangle{BCD}$. Hence, $ \frac{2}{BC}\equal{}\frac{BC}{13}$. Simplifying, $ BC\equal{}\sqrt{26}$. By Pythagorean Theorem, $ BD^2\plus{}2^2\equal{}\sqrt{26}^2\implies BD\equal{}\boxed{\sqrt{22}}$[/hide]

how exactly did you get $ \sqrt{21}$ using the pythagorean theorem?

---

also to get an image follow the directions on the blue bar above the post text area when you hover your cursor over the Img button
or add an attachment by printing the screen and saving it to a file

I think the problem is flawed or something, because BC, if it is a right triangle, would be $ \frac{120}{13}$.

Anyways, here's the pic.

[asy]draw ((0,5)--(12,5)--(12,0)--cycle);
draw ((10.3076923, 0.769230)--(12,5));[/asy]

It's a 5-12-13 right triangle, and BC is the altitude from the right angle.
B is the right angle.

D is the point beneath the right angle.

Some simple math tells us that the altitude is $ \frac{120}{13}$.
Also, in the original picture they marked angles C and B with right angle signs.

you cant assume its a 5-12-13 right triangle !!
in fact its not
(edit: added 5-12-13)
heres why
assume it is
the length of BD would be 120/13 BUT the length of DC would be 25/13 which is not 2

It CAN be a right triangle, but it's just not a 5-12-13 triangle. (In fact, it's a $ \sqrt{26}$-$ \sqrt{143}$-$ 13$ right triangle :D)

Yes, I figured that out a while later... It's a $ \sqrt {26}$, $ \sqrt {143}$, $ 13$ triangle like Yongyi said, and the answer IS $ \sqrt {26 \minus{} 2^2} \equal{} \boxed{\sqrt {22}}$.

I stupidly assumed it was a 5-12-13 right triangle because it looked like one... I have no place on AoPS.  :o

Meh...I just used the fact that the altitude of a right triangle is the geometric mean of the product of the lengths of the two segments it breaks the base into...=P
Easily proven with similar triangles...


Cory Lidle, a pitcher for the New York Yankees, was killed in a plane crash earlier today. He was learning how to fly, when he crashed into a 50 story apartment building. Both he and his instructor were the only casualties. :(  :(  :(

http://mlb.mlb.com/NASApp/mlb/news/article.jsp?ymd=20061011&content_id=1708962&vkey=news_mlb&fext=.jsp&c_id=mlb

Yes, it is karma.

But still, it is sad.

[quote="mustafa"]Cory Lidle, a pitcher for the New York Yankees, was killed in a plane crash earlier today. He was learning how to fly, when he crashed into a 50 story apartment building. Both he and his instructor were the only casualties. :(  :(  :(
[/quote]
whoa. what happened to the peopoe in the buildings

I don't know, it seems strange that nobody else died. Maybe nobody was home that day.

I'm just surprised a plane could get so close to New York without raising any alarms after what happened (9/11)

There are still plane flights to and from NYC, they are just a bit more closely monitered I would guess. Cory Lidle was registered, people knew that he flew and that he was flying that day.

It is very sad. I'm a diehard redsox fan, but when something like this happens, it really doesn't matter what team the death ocurred in. I even wore a NY Yankees cap the day he died t my school.  :(  :(


Let $ T$ be a point outside a circle. From $ T$, we draw two secants. The first secant intersects teh circle in two points $ V$ and $ R$ such taht $ V$ is between $ R$ and $ T$. The second secant intersects the circle in two points $ S$ and $ U$ such taht $ S$ is between $ U$ and $ T$. If the measure of $ \angle RTU\equal{}36$, the measure of $ \angle TRS\equal{}30$, $ RT\equal{}UT$, and $ VU$ and $ RS$ intersect at $ Q$, what is the measure of $ \angle RQV$?

[hide]The Two triangles RQV and UQS are obviously congruent, so $ \angle VUS\equal{}\angle SRV$, or just use inscribed angles.  So since the angles of a triangle add up to 180 degrees, angle $ RST\equal{}114^{\circ}$.

From here, we can find the angle with relative simplicity, giving us $ 84^{\circ}$. Hopefully.[/hide]

I have an alternate solution.

Consider that inscribed angle R is half the degree measure of minor arc VS.  Therefore VS=60 degrees.  We are given that angle T is 36 degrees.  Also, we know that (1/2)(RU-VS)=T, and with substitution, one can solve quickly for RU (132 degrees).  Adding up minor arcs VS and RU, we can subtract it from 360 degrees (the entire circle).  That means that the sum of minor arcs RV and US (the parts of the circle that are not RU and VS) must be 168 degrees.  Half of this is 84 degrees, which is the measure of angle RQV, as desired.

You have to know some geometry facts about inscribed angles, chord-chord angles, and angles with vertices outside a circle with sides intersecting that same circle.


I read this really cool book called Pendragon.  I think it is better than harry potter.  Has anyone read Pendragon?

I've read four or five of the books. They're okay, but they seem to be more for younger children. Not even comparable to Harry Potter in my opinion...they aren't even in my long list of books that are almost as good as Harry Potter.

i've heard of it in a matter of fact the person naxt to me on the bus reads it!

I read the first one if you want to know what it's about click below
[hide]basically it's this guy called Bobby Pendragon who lives a normal life.Then his uncle asks for his help. Then he goes to the planet Denduron. HE must rescue people from the half evil Mallo. It's kinda sci-fi and adventure plus a little fantasy. But just a little  ;)  ;) [/hide]

YEah, I guess you are right. And by the way everyone I made a mistake in my previous post. Its MALLOS not mallo. I just couldn't remember his name.


Thank you!
But I think it is not easy to prove
$f(a,b,c,d) \le f(\frac{{a+b}}{2},\frac{{a+b}}{2},c,d)$
can you solve it in more detail?

Two solution for you, Huyen Vu. The first one uses AC-method in VASC's book (you can find it in pages showing AC-method). The second one is in Hungkhtn's book, solution by a more elementary one: Chebysev  inequality. It seems to be so hard to prove it by the way you say.

[quote="me@home"][quote="fermat3"]ther is a generalisation for your problem see here [url=http://www.imo.org.yu/othercomp/Journ/ineq.pdf]www.imo.org.yu/othercomp/Journ/ineq.pdf[/url]. problem 2389[/quote]
Sorry, I scrolled through the entire thing but can't find number 2389!?[/quote]it's on pg.93 at the bottom

Thank you, also does anyone know where to purchase Vasc or Hungkhtn's book or is it only at the IMO?

See also http://www.mathlinks.ro/Forum/viewtopic.php?t=18627 for this problem.
I guess several people on this forum would be more interested in Hungkhtn's proof rather than in the mention that some unavaliable book contains it.

  darij


The expresion \[ 1*2*3*4*5*6*7*8*9*10 \] , if we can replace the operator $*$ with $-$ or $+$, how many way so that the sum is equal to $29$ !

if they are all positive, it is 55.  By changing a + to -, we reduce the sum by twice the number.

So we want a set of numbers whose sum is $\frac{55-29}{2} = 13$.

Suppose the set contains largest element a 9.  Then we can use 4, or (3,1).
Suppose .... for 8.  Then we can use 5, (4,1), (3,2).
Suppose .... for 7.  Then we can use 6, (5,1), (4,2), (1,2,3).
Suppose .... for 6.  Then we can use (5,2), (4,3), (4,2,1).
Suppose .... for 5.  Then we can use (5,4,3,1).


inequalities
geometry
circumcircle
triangle inequality
geometry proposed

[i]Show that in the acute triangle there's always a pair of sides whose sum is greater than the sum of the diameteres of its incircle and circumcircle[/i]

Sincerely
T.H

[b]Lemma:[/b] Point $ M$ lies inside $ \triangle ABC$ and $a=\min(a,b,c).$ Then we have $ b \plus{} c > MA \plus{} MB \plus{} MC.$

Draw the parallel to $ BC$ passing through $ M$ intersecting $ AC$ and $ AB$ at $ X,Y,$ respectively. Draw the altitude $ AH$  and w.l.o.g assume that $ M$ lies inside $ \triangle AHB.$ We have that $ YA > MA \ (1)$ and since $ AC > CB$ $ \Longrightarrow$ $ AX > XY,$ due to $ \triangle ABC \sim \triangle AXY.$  Thus $ AX \plus{} XC \equal{} b > XY \plus{} XC \ (2)$

By triangle inequality $ MX \plus{} XC > MC \ , \ MY \plus{} YB > MB$

$\Longrightarrow  CX \plus{} XY \plus{} YB > MB \plus{} MC \ (3)$

Adding inequalities $ (1),(2),(3)$ together yields

$ b \plus{} YA \plus{} YB \plus{} XC \plus{} XY > MA \plus{} MB \plus{} MC \plus{} XY \plus{} XC$

$ \Longrightarrow b \plus{} c > MA \plus{} MB \plus{} MC$

The proof is analogous when $ M$ either lies inside $ \triangle AHC$ or on the altitude $ AH.$
____________________________________________

Assume that $ CA \ge AB \ge BC.$ Using the above result when $ M$ is identical to the orthocenter $ H$ of the acute $ ABC,$ we get $b \plus{} c > HA \plus{} HB \plus{} HC \equal{} 2(R \plus{} r).$


calculus
integration
geometry
real analysis
real analysis unsolved

http://www.expertbee.com is a place where we match people who have problems with people who can provide answers.

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ymax=115
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calculus
algebra
polynomial
floor function
limit
function
LaTeX

I am working through Larson's Problem-Solving Through Problems and Ch 1.11 seems to be giving me problems. The concept of extreme cases makes sense, and I can read and understand the example proofs, but I can't seem to apply it to these problems: (one of them is definetly calculus, and the other isnt, but I didnt really know where to post this. Is there a better forum for these sort of questions?)

1.11.5 Let $ f(x)$ be a polynomial of degree $ n$ with real coefficients such that $ f(x) \geq 0$ for any $ x$. Show that $ f(x) \plus{} f'(x) \plus{} \dots \plus{} f^{(n)}(x) \geq 0$ for all $ x$

[hide="my ideas"] I thought maybe you could do something like take the lowest $ n$ such that this isnt true, and then get a $ n \minus{} 2$ polynomial. (-2 because the polynomials should be of even degree to be always positive) but i couldnt figure out how to do that. [/hide]
1.11.7 Show that there exists a rational number $ c/d$ with $ d < 100$ such that $ \left\lfloor k\frac cd\right\rfloor \equal{} \left\lfloor k\frac {73}{100}\right\rfloor$ for $ k \equal{} 1,2,\dots$ Where $ \lfloor x\rfloor$ is the greatest integer less than $ x$

This one has me completely stuck. Can anyone give me hints, solutions, or maybe even a better way to go about thinking about these problems. They always seem like the idea was just a lucky thing to pick. Is there some method to knowing when and how to use extreme cases?

Let $ g(x) = e^{ - x}\left(f(x) + f'(x) + f''(x) + \cdots + f^{(n)}(x)\right).$

Then, $ g'(x) = - e^{ - x}f(x) < 0$ for all $ x,$ and $ \lim_{x\to\infty}g(x) = 0.$

Can you see what you can get from a decreasing function that goes to zero at infinity?

$ \text{\LaTeX}$ tip: use \lfloor and \rfloor, and to make it bigger, \left\lfloor and \right\rfloor.

$ \lfloor x\rfloor$ or $ \left\lfloor\frac{kc}{d}\right\rfloor.$

[quote="Kent Merryfield"]
Then, $ g'(x) \equal{} \minus{} e^{ \minus{} x}f(x) < 0$ for all $ x,$ and $ \lim_{x\to\infty}g(x) \equal{} 0.$

Can you see what you can get from a decreasing function that goes to zero at infinity?
[/quote]

That makes sense, but can you prove it using a "consider extreme cases" sort of method? I dont think I have a good grasp on how to come up with solutions of this sort.

(by consider extreme cases, i mean a proof of the sorts:

we want to prove the statement P(n) for all n, suppose that P(n) is false, then for some reason there must be a minimal (maximal) n such that P(n) is false, then we use this false P(n) to construct a P(n-1) (or n+1, depending) false example, contradicting our minimality (maximality) of n)

let $ g(x)\equal{}f(x)\plus{}f'(x)\plus{}\cdots\plus{}f^{(n)}(x)$. then prove that $ g$ has a minimum. then look at $ g'(y)$ for properly chosen $ y$.

Well since you established that. and $ f(x) \geq 0$ for all $ x$, $ g(y) \geq 0$.

Ohh I see it now. What about the other one

[quote="skylarth"]
1.11.7 Show that there exists a rational number $ c/d$ with $ d < 100$ such that $ \left\lfloor k\frac cd\right\rfloor \equal{} \left\lfloor k\frac {73}{100}\right\rfloor$ for $ k \equal{} 1,2,\dots$ Where $ \lfloor x\rfloor$ is the greatest integer less than $ x$[/quote]

[quote="skylarth"]$ n$ has to be even, so $ g'$ is of odd degree, so it has a zero. so it has a minimum. let $ y$ be where it occurs, so $ g'(y) \equal{} 0$ there. but, $ g'(x) \equal{} g(x) \minus{} f(x)$, so $ g(y) \equal{} f(y)$ there. Is this what you mean? I still dont see what to do next.[/quote]
well assume that $ g(x) < 0$ for at least one $ x$, call this point $ x_0$, so $ g'(x_0) \equal{} 0$. Now $ g'(x) \equal{} g(x) \minus{} f(x)$ so $ g'(x_0) \equal{} g(x_0) \minus{} f(x_0) < 0$ which contradicts $ g$ having a minimum at $ x_0$.

yes, but nobody has been able to do it with an "extreme case" sort of method. anyways, what about the other one? can anyone do that one? i still am having difficulty with it

the 'extreme value' part comes from considering the minimum (extremum) of $ g(x)\triangleq \sum_{0\leq j\leq n} f^{(j)}(x)$ (which is easily seen to exist, since $ f\geq 0$ implies that $ f$ has even degree, and $ g$ has the same degree as $ f$ and leading coefficient as $ f$).  if the minimum of $ g$ is attained at $ t$, then $ g'(t) \equal{} 0$, so $ g(t) \equal{} g(t) \minus{} g'(t) \equal{} f(t)\geq 0$, as desired.

this gets around an argument by contradiction, but it's essentially the same thing.


[quote="neelnal"]Does anyone know a quick way to find base numbers such as:

 145 in base 3

How can I do that quickly?

A proof or some sort of explanation would be appreciated.

Also, I do not understand the concept of "mod". I think it is called modulant in full. How to find this easily and what are the applications of this? An simple example problem and/or proof would be nice. :D[/quote]
I'm not entirely sure about the base conversion, but I can explain modulos.

$10 \bmod 3=x$

$x$ is the remainder when 10 is divided by 3. In this case, it is 1.

$100 \bmod 50=0$

$25 \bmod 7=4$

[quote="neelnal"]Does anyone know a quick way to find base numbers such as:

 145 in base 3

How can I do that quickly?

A proof or some sort of explanation would be appreciated.

Also, I do not understand the concept of "mod". I think it is called modulant in full. How to find this easily and what are the applications of this? An simple example problem and/or proof would be nice. :D[/quote]
To find x in base y quickly, you need to find the largest power of y that is less than x. For example for 145 in base 3 the largest power of three is 81. Only one 81 can fit because 2 81s are too big (162). So the first digit is 1. 145-81=64. The next lowest power of 3 is 27. 27 fits twice (54). So the second digit is a 2. 64-54=10. 9 fits in once. So the third digit is 1. 10-9=1. 3 fits 0 times. So the fourth digit is three. 1 is the leftover making 1 the last digit.
Thus 145 base 3=12101

A number x in modulo y is the remainder when x is divided by y. 84 mod 5 is equal to 4.
An example is find the remainder when 1+17+177+1777+...+1777777777 is divided by 8. You may notice that these numbers are all equal to 1 mod 8. There are 10 numbers. 10 mod 8 is 2. So the answer is two.

[quote="bpms"]3 fits 0 times. So the fourth digit is three. [/quote]
You mean zero, right? :wink:

[quote="i_like_pie"][quote="bpms"]You may notice that these numbers are all equal to 1 mod 8.[/quote]
Does that mean the remainder is 1 after mod 8? Or does that mean it's 8?[/quote]
That means when the numbers are divided by 8 they have a remainder of 1

[quote="mtms5467"][quote="i_like_pie"][quote="bpms"]You may notice that these numbers are all equal to 1 mod 8.[/quote]
Does that mean the remainder is 1 after mod 8? Or does that mean it's 8?[/quote]
That means when the numbers are divided by 8 they have a remainder of 1[/quote]
Ok. Thanks. :)

[quote="i_like_pie"][/quote][quote="mtms5467"][quote="i_like_pie"][quote="bpms"]You may notice that these numbers are all equal to 1 mod 8.[/quote]
Does that mean the remainder is 1 after mod 8? Or does that mean it's 8?[/quote]
That means when the numbers are divided by 8 they have a remainder of 1[/quote][quote="i_like_pie"]
Ok. Thanks. :)[/quote]
How could the remainder of x/8 be 8? It would be 0 if it was in that case.

[quote="bpms"][/quote][quote="i_like_pie"][/quote][quote="mtms5467"][quote="i_like_pie"][quote="bpms"]You may notice that these numbers are all equal to 1 mod 8.[/quote]
Does that mean the remainder is 1 after mod 8? Or does that mean it's 8?[/quote]
That means when the numbers are divided by 8 they have a remainder of 1[/quote][quote="i_like_pie"]
Ok. Thanks. :)[/quote][quote="bpms"]
How could the remainder of x/8 be 8? It would be 0 if it was in that case.[/quote]
$1 \bmod 8 = 8.$

[quote="i_like_pie"][/quote][quote="bpms"][/quote][quote="i_like_pie"][/quote][quote="mtms5467"][quote="i_like_pie"][quote="bpms"]You may notice that these numbers are all equal to 1 mod 8.[/quote]
Does that mean the remainder is 1 after mod 8? Or does that mean it's 8?[/quote]
That means when the numbers are divided by 8 they have a remainder of 1[/quote][quote="i_like_pie"]
Ok. Thanks. :)[/quote][quote="bpms"]
How could the remainder of x/8 be 8? It would be 0 if it was in that case.[/quote][quote="i_like_pie"]
$1 \bmod 8 = 8.$[/quote]
Oh. Sorry for the confusion...

A whole lot of info about mods can be found [url=http://www.artofproblemsolving.com/Wiki/index.php/Modular_arithmetic/Introduction]here[/url]. Our [url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?page_id=10]Introduction to Number Theory[/url] book is a great way to learn about mods and base numbers, too.

I looked up mods (can be found in AoPS volume 1 chapter 5) and it explained it in full detail, and some proofs from mods. In my own words, If you have X (mod y) that means X+dy where d is any integer and so is y and so is X. For example, 3 (mod 5) is congruent (not equal) to 8, -2, 13, -7, etc.  :wink:


Generate the explicit form of the recursion $\ R_{n+1}=(R_n)^2 + C$ , where C is an arbitrary constant.

Is it even possible to derive an explicit expression for non linear recursion?
I have not seen any good text on solving such beasts.

Me neither and I am unsure on how, thats why I posted it on the forum actually.

I don't think you can for the general case (the c = 0 case is, of course, trivial).  Most nonlinear recurrence equations, even relatively simple ones such as the logistic equation, don't have a closed-form solution except in very specific cases (and, just as an interesting note, I plugged the recurrence equation into Mathematica, and it couldn't give any closed-form solution).


modular arithmetic
number theory
divisibility tests

I came across this problem and had to brute force to get it (actually I got it wrong!).

List all 3-digit numbers $abc$ for which the 6-digit number 579$abc$ is divisible by 5,7, and 9.

I know the rules for 5 and 9 but I have no clue for 7. Anyone know if there is any?  :?

I know not a rule for 7, but the simplest that I see of going about this (tho it might be the way you tried, but it shouldn't be that bad) is to note that if a number is divisible by 5, 7 and 9, then it must be divisible by $3^25^17^1=315$

$\frac{580000}{315} \approx 1841.2$, so $1841 \times 315 = 579915$

The rest is subtraction, $579915,579600,579285$.  Trying to work through divisibility rules seems like it would end up a lot more complicated.

A known divisilibity criterion (and a good exercice) for $7$ is the $a-2 \cdot b$ rules
I.e. $a_1...a_n$ is divisible by $7$ iff $a_1...a_{n-1} - 2 \cdot a_n$ is divisible by $7$

But here $579abc = abc + 2 = 2a + 3b + c \ mod\ 7$

[quote="prototypex24"]A method for determining divisibility for 7 goes like this:

take the last digit of the number, double it and subtract in from the rest. If the result is divisible by 7, so is the initial number[/quote]

and that is just what tutu mentioned  :P

[quote]and that is just what tutu mentioned[/quote]

I've just noticed  :P , only in simpler terms

another method would be multiplying the units digit by 1, the tens digit by 3, the hundreds digit by 2, the thousands digit by -1, the ten thousands digit by -2 and the hundred thousands digit by -3.

It gives:
$c+3b+2a\equiv 5\pmod 7$

Oh okay.

Thanks a lot!

That's what the solution did (to djshowdown2) but I was wondering if there was a quicker way.

Here would be another rule for 7:

Since $1001=7*11*13$, $10^3\equiv -1\pmod{7}$, you can see if a large number is divisible by 7 by taking the "alternating sum" of the periods. (i.e if 579abc is divisible by 7, then so is abc-579). The periods go from the right, so using it on abcde would give cde-ab as the alternating sum. (This can also be used for divisibility by 13 for the same reason.)

Of course, this will only work if there are 4 or more digits. For three digits (this one's only for divisibility by 7), multiply the hundreds digit by 2 and add to the two digit number with the first digit dropped (i.e. 329 is divisible by 7 because so is (2*3)+29=35). This one works because $100\equiv 2\pmod{7}$.

Wasn't that what was written before?  :P

The notation $a\equiv b\pmod{m}$ simply means $a-b$ is a multiple of $m$.

Many properties of equations can also be applied to congruences: $a\equiv c$, $b\equiv d\Rightarrow$

$a+b\equiv c+d$, $a-b\equiv c-d$, $ab\equiv cd$, $a^n\equiv b^n$

(That fourth one plays a big part in divisibility tricks: just expand the number out in terms of powers of 10 (or whatever base).)


A tangent TP is drawn to a fixed circle from a point T on a fixed line passing through the centre of the circle.

A point Q is taken on the normal at P such that PQ =PT.

Find the locus of Q

Hi!

I know that the normal from a point on a circle is a line perpendicular to the tangent

Here P is not on the circle. So, what is its normal line?

[quote]
Here P is not on the circle. So, what is its normal line?[/quote]

To my understanding, P is the point of tangency.

OK :)

I thought that $T$ was on the circle

Here is my solution. Is it correct or am I missing something???

Let $Oxy$ the coordinate system and $(O,1)$ the fixed circle.

Let $x'x$ be the fixed line on which $T$ is moving.

Since there are [color=blue]two[/color] tangent segments $TP,TP'$ from $T$, we have two points instead of $Q$:

$Q(x,y)$ and $Q'(x,-y)$ , where $y > 0$

(when $T$ tends to the point $(1,0)$ we have $Q , Q' \to (1,0)$)

Let's find the polar coordinates $(r,\theta)$ for $Q$.

While $T$ is moving from the point $(1,0)$ to $(\infty,0)$, the angle $\theta = \angle TOP$ is increasing from $0$ and tends to $\frac{\pi}{2}$.

[img]http://i7.tinypic.com/24vtf0n.gif[/img]
$Q$ is on the ray $OP$, so in polar coordinates the angle of $Q$ is $\theta$. What we need is the distance from the origin  $r=OQ$

But $OQ = OP+PQ = 1+PT \ \ \ (1)$

From the right triangle $OPT$ we have $PT = OP \cdot \tan{\theta}\Rightarrow PT = \tan{\theta}$

so we have $(1) \Rightarrow r(\theta) = 1+tan(\theta) ,0 \le \theta < \frac{\pi}{2}$

From here, all are nothing but symmetry

The point $Q'$ is symmetric to $Q$ with respect to $x$ axis. So $Q' = (r, \theta)$ where $r = 1+tan(-\theta)= 1-tan(\theta)$

The problem states that $T$ is moving rather on a line than a ray (but not in the interior of the circle)

So, if we move $T$ from $(-1,0)$ to $(-\infty,0)$ we will find that the two obtained points are symmetric to the other two.

So we have

$r(\theta) = \{\begin{array}{cc}1+tan(\theta) ,&\mbox{ if }\theta \in (0, \frac{\pi}{2}) \\ \\ 1-tan(\theta) ,&\mbox{ if }\theta \in (\frac{\pi}{2}, \pi) \\ \\ 1+tan(\theta) ,&\mbox{ if }\theta \in (\pi,\frac{3\pi}{2}) \\ \\ 1-tan(\theta) ,&\mbox{ if }\theta \in (\frac{3\pi}{2}, 2\pi) \end{array}$

or better, $r(\theta) = 1+|tan(\theta)|$


Suppose you have a rod and you hang a rope over it. Assume there is a coefficient of friction $ \mu$ between the rod and the rope, and that the rod does not rotate. Apply $ T_1$ on one side of the rope and $ T_2$ on the other side. The rope is in contact with the rod over an angle $ \theta$ (the central angle).

How does one derive the formula $ \frac{T_2}{T_1}\equal{}e^{\mu\theta}$?

that should probably be a less than or equal to sign, right ?

so consider an infinitesimal angle $ d\theta$; suppose the tension at the left part of this infinitesimal piece is $ T$ and at the right is $ T \plus{} dT$. Let $ \phi \equal{} \frac {1}{2} d\theta$ and let the midway point of the arc be vertically up. Then the left tension resolves to $ T \langle \minus{} \cos \phi, \minus{} \sin \phi \rangle$ and the right tension to $ (T \plus{} dT) \langle \cos \phi, \minus{} \sin \phi \rangle$; we have some normal force $ N$ that points straight up and some frictional force $ f$ that points, to first-order, horizontally. So it looks like $ N \equal{} (2T \plus{} dT) \sin \phi$ and $ dT \cos \phi \equal{} f \le \mu N \equal{} \mu (2T \plus{} dT) \sin \phi$. Now, to first order, $ \cos \phi \equal{} 1$ and $ \sin \phi \equal{} \phi \equal{} \frac {1}{2} d\theta$ so the above equation reads $ dT \le \mu (2T \plus{} dT) \frac {1}{2} d\theta \equal{} \mu T d\theta$. Integrating yields $ \exp \minus{}\mu \theta \le \frac{T_2}{T_1} \le \exp \mu \theta$.


Who's the mathlinker with most posts??? and with most posts per day???

Click the 'memberlist' button at the top.


algorithm
algebra
polynomial
calculus
number theory
greatest common divisor
modular arithmetic

Hi,

I'm sorry to open a new thread but it is necessary because I found something interesting.
I want your thoughts.
Thank you!

n=p*q (p et q odd primes)

Theorem 1

2^((p-1)*(q-1)) mod pq = 1

Let

(p-1)*(q-1)= pq + 1 -(p+q)

So (2^(pq+1) mod pq) * (2^(-p-q) mod pq) = 1

The first part a=(2^(pq+1) mod pq) is easily computable
The secund one is unknown x=(2^(-p-q) mod pq) but the equation is easily solvable

ax mod n = 1

It is a linear congruence. So the solution is known.

The factorization is complete when x is identified.

Any comment?

Hi,

Here is an example.
I know that you are going to ask me about how to solve y

Any comment?

Hi,

It seems in another forum that it is not easy to solve :

a*(2^y mod n) = b

What if I have a way to solve this equation?

I just want to know if there is no other problem than this one.

Hi,

I want just to know what is wrong in my method.
Can someone explain step by step where are the errors?

Thank you for any explanation.

Good bye and good luck to all.

Thank you anyway.

Hi,

I came back because I still do not understand why it is so hard to solve :

(2^y)*a mod n= b

If we replace 2^y = x

The equation will become :

a*x mod n=b

We know how to solve it using extended Euclidean algorithm.

[u]This algorithm runs in time O(log(m)^2), assuming |a|<m, and is generally more efficient than exponentiation.[/u]

from wikipedia http://en.wikipedia.org/wiki/Modular_multiplicative_inverse

So where is the problem?
Is it hard when a number get larger?
We have 2 steps to factorize a semi-prime :
- exponentiation
- solution of ax mod n=b

Can someone compute the complexity of my algorithm (read above)?
How long it could take if we try to factorize a semi-prime of 1000 digits?
Thank you for any answer.

I have found a very good table of complexity

http://en.wikipedia.org/wiki/Computational_complexity_of_mathematical_operations

Exponentiation by squaring O(n M(k))
Euclidean algorithm O(n^2)

Is my method a polynomial time algorithm?

No one is answering to my questions.
I checked a lot of numbers my method is working.
I do not know how to prove it.
I suppose that my statements above are right and correct if no one has proved it wrong.

I googled a little bit on internet to find the complexity of the extended Euclidean algorithm.
Here is what I have found in french

[i]L'algorithme d'Euclide est donc très efficace : si l'on dispose d'une bibliothèque d'opérations sur les entiers de taille quelconque (comme en Maple ; on parle alors de calculs en multiprécision) calculer le PGCD de nombres de 100 chiffres ne demande que quelques centaines d'opérations. [/i]

http://www.labri.fr/perso/betrema/deug/poly/euclide.html#beziter

I translate :
... to compute GCD of 100 digit numbers require only few hundreds of operations.

So my algorithm is quite simple.
3 steps are needed :
- computing by modular exponentiation some values
- computing square root of the number n plus multiplication by 2
- solving ax mod n= b where x is equal to 2^y

I do not understand why no one is talking the big advantage of my algorithm.
[u]You can solve a semi-prime RSA like in less than 10 minutes with appropriate tools (multiprecision and so on)!!![/u]
Some one told me that it is like solving discrete logarithm. [u]It has nothing to do with discrete logarithm.[/u]
I replaced 2^y by a difference z-x where I can choose x such I can eliminate the number a to find directly the solution.
I found 4096 and then it is easy to cumpute y by dividing by 2 and counting so y=12.
I did it with other numbers little bit bigger than 26697. It works!

I need someone to explain me why.
I do not have big knowledge in number theory that is why I'm asking for your help.
Thank you for any useful comment.

[quote="Nukular"]I want to avoid sounding as rude as possible here, but could you *clearly* state what your question is? I read through
your posts and it's not clear to me which things you are asserting as facts, and which things you would like to know.
I apologize, but it is very hard to read your posts -- probably a language barrier thing, but please try to be clear.[/quote]

Thank you for your answer.
I prefer that you be rude than silent.
Let us start from the beginning.

First statement :
n=p*q where p and q are prime.
n is a semi-prime.
s= (p-1)*(q-1)
2^s mod n = 1

Example n=91
n=7*13
s=(7-1)*(13-1)=6*12=72

2^72 mod 91=1

Is this statement true?

An answer and then I will continue.

If $ p$ and $ q$ are distinct odd primes, certainly: the result follows directly from
Fermat's Little Theorem; $ 2^{(p\minus{}1)(q\minus{}1)} \equiv 1 \pmod{p}$, and $ 2^{(p\minus{}1)(q\minus{}1)} \equiv 1 \pmod{q}$.

If the last statement is true then

s=(p-1)*(q-1)
 = pq - p - q +1
 = pq -(p+q)+1
 = pq +1 - (p+q)

2^s = (2^(pq+1))/((2^(p+q))

2^s mod n = 1

2^(pq+1)mod n = 2^(p+q) mod n

Example n=91

2^92 mod 91 = 74
2^20 mod 91 = 74

Is the secund statement true?

You've made no claim above, just the definitions of $ t$ and $ y$; it's clear $ y > 0$ by AM-GM.
But if that's what you meant to do, OK.

So now we have to solve a linear congruence

ax mod n=b

where
a=64
x=2^y
b=74

If we can find x which is equal to 4 we obtain by dividing by 2 y=2

p+q = 18+2 = 20

it is done!

But how are you computing $ b$?

EDIT: I assume via $ b \equal{} 2^{n \plus{} 1} \pmod{n}$, no?

EDIT 2: Sure, but solving $ 2^y \equiv x \pmod{n}$ is hard.

[quote="Nukular"]But how are you computing $ b$?[/quote]

b=2^(pq+1)

I told you that above.

Edit : Yes. We know b by computing 2^(n+1)

[quote="Nukular"]OK, so how are you computing $ y$? You can find $ x$ and $ b$ in about $ O(\log n)$ time, but how do you find $ y$ such that $ 2^y \equiv x \pmod{n}$?[/quote]

That is a good question.

There is only one solution for x.
I tried a lot of numbers.
I can not explain it nor prove it. That is why I need help.
Try it first one some big numbers to see if it works. I do not have the tools for big numbers (100 digits and more).
Then we can find why.

I'll give you one better: if you can solve equations of the form $ 2^y \equiv x$, then why not just solve $ 2^{p\plus{}q} \equiv b \pmod{n}$ directly? Why do all that stuff with $ t$ and $ y$?

This problem is generally very difficult to solve; this is the "discrete logarithm" you mentioned earlier. I don't think this is going to reduce
to anything nice.

There are many $ y$ that will work; your value of $ y$ is only up
to a multiple of $ \mathrm{ord}_2(n)$. If you could compute this,
you would have the factorization of $ n$ immediately anyway.

[quote="zeb"]You did indeed find a way to compute $ x$! And finding $ y$ would indeed give you the factorization! I am amazed and would like to give you all of my money! But wait...

How do you find $ y$ once you know $ x$? This is the "and then a miracle occurs" step in your method. Let me give you an example (I actually took the time to find some smallish primes on the internet):

$ pq \equal{} 8876044532898802067$.

$ 2^{pq \plus{} 1} \equiv 4070778346170831458 \mod 8876044532898802067.$ (this was annoying to compute, but didn't take me too long with a calculator)

So $ 4070778346170831458x \equiv 1 \mod 8876044532898802067$. Oh crap, I have to run the Euclidean algorithm... time to download a new computer program...

It tells me $ 2^y \equiv x \equiv 3225243598127278184 \mod 8876044532898802067$. Now tell me... what's $ y$? Take as long as you like. You can download some software, write some software, whatever. Just tell me what $ y$ is so I can finish factoring this. I'm really curious!!

Edit: the program I downloaded is called "genius". You can find it on the interwebs.[/quote]

y=6

Check it

Yes, up to a multiple of $ n$; multiplicative inverses are unique when they exist!

Thank you to all of you for your help.
Now I know what I'm going to do to solve the question.
I have an idea.

I did not check the results above given by Zeb because I was really tired.
Anyway if I find another way to express the factorization of any semi-prime as linear congruence then it will be ok.
For some kind of semi-primes my method as presented above works.
There are conditions to set up.

You should look up Peter Shor's factoring algorithm on the interwebs. The idea is very similar, except instead of using $ 2$ he uses a random number $ a$, and to solve $ a^y\equiv x \mod pq$ he uses a Quantum Computer! This should give you an idea of how hard it is to solve congruences like $ 2^y \equiv x$.

Most of the other factoring techniques I've seen either try to solve congruences such as $ x^2\minus{}y^2\equiv 0 \mod pq$, or try to find the period of some function, such as doubling a number mod $ pq$, squaring a number mod $ pq$, or doubling a point in an elliptic curve. Some of them work very well when $ p\minus{}1$ factors into small primes, or when $ p\plus{}1$ factors into small primes. Really, you should look a lot of these up - you'll see a lot of your ideas are already being used (and they are mostly simple enough for a highschool student to understand them).

[quote="zeb"]You should look up Peter Shor's factoring algorithm on the interwebs. The idea is very similar, except instead of using $ 2$ he uses a random number $ a$, and to solve $ a^y\equiv x \mod pq$ he uses a Quantum Computer! This should give you an idea of how hard it is to solve congruences like $ 2^y \equiv x$.

Most of the other factoring techniques I've seen either try to solve congruences such as $ x^2 \minus{} y^2\equiv 0 \mod pq$, or try to find the period of some function, such as doubling a number mod $ pq$, squaring a number mod $ pq$, or doubling a point in an elliptic curve. Some of them work very well when $ p \minus{} 1$ factors into small primes, or when $ p \plus{} 1$ factors into small primes. Really, you should look a lot of these up - you'll see a lot of your ideas are already being used (and they are mostly simple enough for a highschool student to understand them).[/quote]

I'm trying to find some direct way to solve the problem without using random method.
I still believe that [u]the ideas are more important [/u]than the tools.
I have nothing to loose by debating my ideas maybe someone who is better equiped will finally find the solution. It does not matter for me who find it.
Anyway here is my idea about solving 2^y mod n = b
I do not know where it leads.

2^y = b mod n

We have to solve this equation.
My idea is as follows :
If we assume that for some value v we have

y+v prime

then

2^(y+v-1) mod (y+v) = 1

2^(y+v-1) = k*(y+v) + 1

2^y = (k*(y+v) + 1)/(2^(v-1))

v can be used as parameter

(k*(y+v) + 1)/(2^(v-1))mod n = b

(k*(y+v) + 1) mod n = b * (2^(v-1))

k*(y+v)mod n = (b * (2^(v-1))) - 1

We will have 2 unknown values y and k
and one parameter v

By replacing v by some value assuming that v is the right value giving y+v prime
we will have to solve an equation

x mod n = d

where

d=((b * (2^(v-1))) - 1) mod n
z=y+v
x=kz

We know yet one of the value of y (in terms of modulo n) which is n+1

If we replace it we could find k

So the equation will be

k*(1+v) mod n = (b * (2^(v-1))) - 1

or k*(1+v) mod n = d

It will be easy to find k by computation

then we can rewrite the equation with only one unknown value which is y

Any comment?

So now you're trying to solve an equation like $ v \plus{} \mbox{stuff} \equiv \mbox{blah}*2^v \mod n$. I'd say you've only made the problem harder.

Edit: Or wait, I misunderstood. You switched gears halfway through and started trying to find $ k$. You seem to think $ k$ will be the same for all values of $ y$...

Whatever. If you can get it to work to factor the number I gave above, I'll be more interested. Now it just looks like you are throwing out random ideas hoping something will work.

I have found a [u]general way[/u] to solve 2^y mod n = b

I think that it is workable.
I will post in 2 or 3 days.

I'm very sorry I did a miscalculation.
So forget it!

My idea was wrong.
It works for some cases because of the specificity of those numbers.

Now I'm hundred per cent sure that I have found the solution.
I need the help of one of the moderator to write it correctly with latex and an english easy to read.
Please if someone of you is not busy let me know.
I will send him step by step my algorithm using private message.
Thank you for any help.


Source: Massachusetts Mathematics League March 2003

One of the terms in the expansion of $(a + b)^n$ is $_{-7x+6y-7}C_{2x+5y}a^{4x+3y}b^{2x+5y}$. If the number of this term in the expansion is nine more than the exponent of $a,$ calculate the third term in the expansion in simplified form.

[hide]
The kth term of a binomial expansion $(a+b)^n$ is given by $\displaystyle\binom{n}{k-1}a^{n-k+1}b^{k-1}$
Our particular term is $\binom{-7x+6y-7}{2x+5y}a^{4x+3y}b^{2x+5y}$, so $n=-7x+6y-7$. The exponent of $a$ and the exponent of $b$ always add up to $n$, so
$\begin{equation*} -7x+6y-7=(4x+3y)+(2x+5y) \end{equation*}$ (1)
[quote="weatherbottle"]... the number of this term in the expansion is nine more than the exponent of $a$ ...[/quote]
so $k=(n-k+1)+9$
$\begin{equation*} (2x+5y)+1=(4x+3y)+9 \end{equation*}$ (2)
Combining (1) and (2), we get $(-1,3)$ for our solution. Substitute these values into $n=-7x+6y-7$, get $n=18$.
The third term of $(a+b)^{18}$ is given by $\displaystyle\binom{18}{2}a^{16}b^{2}=153a^{16}b^{2}$

[/hide]


When you try to type a title in ALL CAPS, it gives you this message:

[quote]All the letters in your topic title are uppercase (A, B, C, ...). Please make some of them lowercase (a, b, c, ...)[/quote]

I'm guessing it's to prevent SPAM or SHOUTING? But I think it does more harm than good, because abbreviations are often used, like "USAMO 2000 #1," and you have to find a way to work around the restriction.

This has caused many inconveniences including:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=126788017&t=238988[/url]
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=126788017&t=216849[/url]
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=126788017&t=131461[/url]

No it's good that way! There are just too many idiots writing uppercase for no reason.

Just add some problem description, something like "USAMO ..." does [b]not[/b] belong to the title but to the source field!

In the links above, I had no idea what any of the acronyms meant; their meanings were not even included in the first post. In fact, the first thread highlighted this confusion, since the first few posts were made to clarify what he was talking about. They would have been much better if the thread maker described what the thread was about.

Rather than find a way to work around the restriction, he should have spent some time thinking about how the thread was presented.

Uppercase Title Restriction is very good. Try finding all threads which should start with only uppercases and look how many threads would we have because of idiotic spammers.

it's quite easy to get around that

say you have "AIME #1 2008"
just put "AIME problem 1, 2008" or something close to that

taking your first link for example, it could say "are you taking OPLET" (i think it was OPLET...)

[quote="007math"]I would say to follow how people post problems onto the Wiki, post the subject of the problem as subject and the competition stuff in the Subject Description.[/quote]

um, what? that's not how one posts problems on the wiki at all.

Don't people go like:

Saturday Evening
[size=75]AMC 10 2005A Problem 6[/size]

no; that's the on the forums for the resource center. wiki pages don't even have equivalents of subject descriptions.


floor function
logarithms
inequalities
function
number theory unsolved
number theory

$ \phi>0>\psi$ - solutions of $ x^2\equal{}x\plus{}1$......
prove that for all $ n$
\[ \left\lfloor \log\frac{\phi^{2n\plus{}1}\minus{}\psi^{2n\plus{}1}}{\sqrt{5}}\right\rfloor\equal{}\left\lfloor\log\frac{\phi^{2n\plus{}1}}{\sqrt{5}}\right\rfloor\]

By the Binet formula the $ k$th Fibonacci number is

$ F_k \: = \: \frac{\phi ^k - \psi^k}{\sqrt{5}}.$

where $ \phi = \frac{1+\sqrt{5}}{2}$ and $ \psi = \frac{1-\sqrt{5}}{2}$. Hence the identity

$ (1) \;\; \left\lfloor \log \frac{\phi^{2n+1}-\psi^{2n+1}}{\sqrt{5}} \right\rfloor \: = \: \left\lfloor\log\frac{\phi^{2n+1}}{\sqrt{5}}\right\rfloor$

is equivalent to

$ (2) \;\; \lfloor \log F_{2n+1} \rfloor \: = \:  \left\lfloor \log \left( \(F_{2n+1} + \frac{\psi^{2n+1}}{\sqrt{5}} \right) \right\rfloor.$

By inspection the identity (2) is correct for $ n \leq 2$. Therefore we may assume that $ n > 2$.
Moreover $ -1 < \psi^{2n+1}/\sqrt{5} < 0$ for all natural numbers $ n$ since $ \psi \approx  -0,62$. Consequently

$ (3) \;\; F_{2n+1} - 1 \: <\: F_{2n+1} + \frac{\psi^{2n+1}}{\sqrt{5}} \: <\: F_{2n+1},$

Next let $ m$ be the numbers of digits in $ F_{2n+1}$, i.e.

$ (4) \;\; 10^{m-1}\: \leq \: F_{2n+1} \: <\: 10^m.$

Next assume $ F_{2n+1} = 10^{m-1}.$ Now $ F_{2n+1} \geq F_7 = 13 > 10$ since $ n>2,$ giving $ m>1$. Hence the only prime factors of $ F_{2n+1}$ are 2 and 5. The smallest Fibonnaci number which divides 10 is $ F_{15} = 610$.  Carmichael's theorem states that for $ k > 12$, the $ k$th Fibonacci number $ F_k$ has at least one prime factor that is not a factor of any earlier Fibonacci number. Since $ 2 = F_3$ and $ 5 = F_5$, Carmichael's theorem implies that all Fibonacci numbers $ F_k>5$ has at least one prime factor other than 2 and 5. Thus the assumption $ F_{2n+1} = 10^{m-1}$ is false. Therefore $ F_{2n+1} \geq   10^{m-1}+1,$ which combined with the inequalites (3) and (4) implies that

$ (5) \;\; 10^{m-1} \: <\: F_{2n+1} + \frac{\psi^{2n+1}}{\sqrt{5}} \: <\: F_{2n+1} \: <\: 10^m.$

Using the logarithm function on (5) we obtain

$ m-1 \: <\:  \log \left( \(F_{2n+1} + \frac{\psi^{2n+1}}{\sqrt{5}} \right) \: <\: \log F_{2n+1}  \: <\: m .$

Consequently

$ \lfloor \log F_{2n+1} \rfloor \: = \:  \left\lfloor \log \left( \(F_{2n+1} + \frac{\psi^{2n+1}}{\sqrt{5}} \right) \right\rfloor \: =\: m-1,$

which in fact is identity (2). [b]q.e.d.[/b]


function
limit
Putnam
real analysis
real analysis solved

Given continious function $f:[0,+\infty){\rightarrow}[0,+\infty)$,such that $\lim_{n{\rightarrow}\infty}{f(nh)}=0(n\in{N}),\forall h>0$.Prove that $\lim_{x{\rightarrow}{+\infty}}{f(x)}=0$.

I don't like to say this all the time, but this has also been discussed before. Unfortunately, I couldn't think of any key words to use, so I couldn't find it. Does anyone remember who the author of that topic was?

Guys,I'm very interested in the solution which is different from mine.

Finally, I've managed to find it: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=17885&highlight=putnam]here[/url]'s the link.

So how about your solution? Does it resemble what you found over there?

The solution proposed by you,is the same as mine:I used one basic idea
if $a<b$ then then there exist $A>0$ such that the set $(A,+\infty)$ is covered by the set $\{{n[a,b]}|n\in{N}\}$.

Can u give a counterexample $(\lim_{x{\rightarrow}\infty}{f(x)}$ doesn`t exist $)$ for a continous functions like the one from the problem but knowing that $\lim_{n{\rightarrow}\infty}{f(nh)}=l_h(n\in{N}),\forall h>0 ?$

But if you find $h_1$ and $h_2$ for which $l_{h_1}\ne l_{h_2}$, then you already have a counterexample. On the other hand, if all $l_h$ are the same (call this common value $l$), then the fact that $\lim_{x\to\infty}f(x)=l$ follows from the problem discussed before applied to $g(x)=f(x)-l$.


At the penny store, there are candies that cost 7 cents and candies that cost 17 cents. If Araz spent exactly 99 cents on these two types of candy, how many total candies did he buy?

$ 99 \minus{} 17 \times 1 \equal{} 82$
$ 99 \minus{} 17 \times 2 \equal{} 65$
$ 99 \minus{} 17 \times 3 \equal{} 48$
$ 99 \minus{} 17 \times 4 \equal{} 31$
$ 99 \minus{} 17 \times 5 \equal{} 14 \equal{} 7 \times 2$
$ 99 \minus{} 17 \times 6 \equal{} \minus{}3$

Thus, $ 5\plus{}2 \equal{} \boxed{7}$.

Alternatively, look at the prices mod 10. They are both 7 cents. Total: 9 mod 10, and 7*7=49=9 mod 10, so 7 is the answer. (Should be logical that 17 candies is too much = 17*7=119)

$c \equiv a \mod b$ if and only if $c-a$ is a multiple of $b$.

For example, $19$ is $9\mod 10$ since $19-9=10$


probability
logarithms
modular arithmetic
number theory unsolved
number theory

(i) Show that a randomly chosen odd number $ N$ has probability $ \frac {2}{\ln{N}}$ of being prime.

(ii) Show that a randomly chosen number $ N$ satisfying $ N\equiv 1\pmod{3}$ has probability $ \frac {3}{2\ln{N}}$ of being prime.


function
integration
logarithms
trigonometry
complex analysis
complex analysis unsolved

Let f: C-->C be an entire function. Show that if

|f(z)|[u]<[/u] 1/(|Re(z)|)

for all z in C, then f is identically the zero function.

What happens if we replace the condition with

|f(z)|[u]<[/u] 1/(|Im(z)|)

for all z in C?

$ u(z) \equal{} \log|f(z)|$ is a [url=http://en.wikipedia.org/wiki/Subharmonic_function]subharmonic function[/url]. Use the sub-mean-value property (here $ iy\in\mathbb R$):
\[ u(iy)\le\frac {1}{2\pi}\int_0^{2\pi}u(iy \plus{} re^{it})dt\le \minus{} \frac {1}{2\pi}\int_0^{2\pi}(\log r \plus{} \log \cos t)dt\to \minus{} \infty
\]
as $ r\to\infty$. Since $ f(iy)\equiv0$, it follows that $ f\equiv 0$.

For the second part: consider $ f(iz)$.


$ a,b,c$ are real numbers and $ a<b$. If for all $ x$, $ ax^{2}\plus{}bx\plus{}c \geq 0$ then find minimum value of $ \frac{a\plus{}b\plus{}c}{b\minus{}a}$.

[quote="mayhem"]$ a,b,c$ are real numbers and $ a < b$. If for all $ x$, $ ax^{2} \plus{} bx \plus{} c \geq 0$ then find minimum value of $ \frac {a \plus{} b \plus{} c}{b \minus{} a}$.[/quote]

$ a<b$--> $ b\minus{}a>0$--->
$ x\equal{}1$--> $ a\plus{}b\plus{}c\geq 0$ ---> $ \frac{a\plus{}b\plus{}c}{b\minus{}a}\geq 0$---> $ min(\frac{a\plus{}b\plus{}c}{b\minus{}a})\equal{}0$

[quote="arcsin1.01"][quote="zaya_yc"][quote="mayhem"]$ a,b,c$ are real numbers and $ a < b$. If for all $ x$, $ ax^{2} \plus{} bx \plus{} c \geq 0$ then find minimum value of $ \frac {a \plus{} b \plus{} c}{b \minus{} a}$.[/quote]

$ a < b$--> $ b \minus{} a > 0$--->
$ x \equal{} 1$--> $ a \plus{} b \plus{} c\geq 0$ ---> $ \frac {a \plus{} b \plus{} c}{b \minus{} a}\geq 0$---> $ min(\frac {a \plus{} b \plus{} c}{b \minus{} a}) \equal{} 0$[/quote]When does the equality (in your answer) hold?[/quote]

zaya_yc must make a mistake.

My solution:
Let min. value attains when $ (a_1,b_1,c_1)$ .If $ a_1x^2 \plus{} b_1x \plus{} c \ge k$ for all $ x$ for a $ k>0$, then $ (a_1,b_1,c_1\minus{}k)$ can give the expression a even smaller value. So min. value attains when $ b^2\equal{}4ac$, the expression becomes $ \frac{(2a\plus{}b)^2}{4a(b\minus{}1)}$. Now let $ k\equal{}b\minus{}a$, it becomes $ \frac{9}{4}*\frac{a}{k}\plus{}\frac{1}{4}*\frac{k}{a}\plus{}\frac{3}{2} \ge 3$ by A.M.-G.M.

So min. value is $ 3$ and it attains when $ \frac{9}{4}*\frac{a}{k}\equal{}\frac{1}{4}*\frac{k}{a}$ , i.e. $ a\equal{}1, b\equal{}4, c\equal{}4$


graph theory
3D geometry
prism
combinatorics
Coloring
IMO
IMO 1979

We consider a prism which has the upper and inferior basis the pentagons: $A_{1}A_{2}A_{3}A_{4}A_{5}$ and $B_{1}B_{2}B_{3}B_{4}B_{5}$. Each of the sides of the two pentagons and the segments $A_{i}B_{j}$ with $i,j=1,\ldots$,5 is colored in red or blue. In every triangle which has all sides colored there exists one red side and one blue side. Prove that all the 10 sides of the two basis are colored in the same color.

Sorry to revive this. The idea of  the solution is from the book A Collection of Problems Suggested for IMO.

[color=#4000BF]Solution:[/color]

Let us prove first that the edges $A_1A_2,A_2A_3,\cdots ,A_5A_1$ are of the same color. Assume the contrary, and let w.l.o.g. $A_1A_2$ be red and $A_2A_3$ be green. Three of the segments $A_2B_l, (l = 1, 2, 3, 4, 5)$, say $A_2B_i,A_2B_j,A_2B_k$,have to be of the same color, let it w.l.o.g. be red. Then $A_1B_i,A_1B_j,A_1B_k$ must be green. At least one of the sides of triangle $B_iB_jB_k$, say $B_iB_j$ ,must be an edge of the prism. Then looking at the triangles $A_1B_iB_j$ and $A_2B_iB_j$ we deduce that $B_iB_j$ can be neither green nor red, which is acontradiction. Hence all five edges of the pentagon $A_1A_2A_3A_4A_5$ have thesame color. Similarly, all five edges of $B_1B_2B_3B_4B_5$ have the same color.We now show that the two colors are the same. Assume otherwise, i.e.,that w.l.o.g. the $A$ edges are painted red and the $B$ edges green. Let us call segments of the form $A_iB_j$ diagonal ($i$ and $j$ may be equal). We now count the diagonal segments by grouping the red segments based on their $A$ point, and the green segments based on their $B$ point. As above, the assumption that three of $A_iB_j$ for fixed $i$ are red leads to a contradiction. Hence at most two diagonal segments out of each $A_i$ may be red, which counts up to at most $10$ red segments. Similarly, at most $10$ diagonal segments can be green. But then we can paint at most $20$ diagonal segments out of $25$, which is a contradiction. Hence all edges in the pentagons $A_1A_2A_3A_4A_5$ and $B_1B_2B_3B_4B_5$ have the same color.

Another way to end the solution (starting from the assumption that $A_1A_2A_3A_4A_5$ is completely red and $B_1B_2B_3B_4B_5$ is completely blue) is to consider $A_1B_j$ for fixed $i$ and $j=1,2,3,4,5$. There must be at least three of the same colour by Pigeonhole. However, if they're red, you get a contradiction (using work from proving $A_1A_2A_3A_4A_5$ is only made up of one colour), and if they're blue, you also get a contradiction (use $\triangle A_1B_xB_y$, where $B_x$ and $B_y$ are two adjacent vertices with $A_1B_x$ and $A_1B_y$ being part of the "at least three of the same colour" group).


[quote="Leon"]Let $a,b,c$ be real positive numbers with $abc=1$. Show that \[\frac{a}{{b^{2}\left({c+1}\right)}}+\frac{b}{{c^{2}\left({a+1}\right)}}+\frac{c}{{a^{2}\left({b+1}\right)}}\geqslant \frac{3}{2}\] [/quote]
Let $a=\frac{x}{z},$ $b=\frac{y}{x}$ and $c=\frac{z}{y},$ where $x>0,$ $y>0$ and $z>0.$
Thence, $\sum_{cyc}\frac{a}{b^{2}(c+1)}=\sum_{cyc}\frac{x^{3}}{yz(y+z)}=\sum_{cyc}\frac{x^{4}}{xyz(y+z)}\geq\frac{(x^{2}+y^{2}+z^{2})^{2}}{2xyz(x+y+z)}.$
Id est, remain to prove that $\frac{(x^{2}+y^{2}+z^{2})^{2}}{2xyz(x+y+z)}\geq\frac{3}{2},$ which obvious.

[quote="arqady"]
Let $a=\frac{x}{z},$ $b=\frac{y}{x}$ and $c=\frac{z}{y},$ where $x>0,$ $y>0$ and $z>0.$
Thence, $\sum_{cyc}\frac{a}{b^{2}(c+1)}=\sum_{cyc}\frac{x^{3}}{yz(y+z)}=\sum_{cyc}\frac{x^{4}}{xyz(y+z)}\geq\frac{(x^{2}+y^{2}+z^{2})^{2}}{2xyz(x+y+z)}.$
Id est, remain to prove that $\frac{(x^{2}+y^{2}+z^{2})^{2}}{2xyz(x+y+z)}\geq\frac{3}{2},$ which obvious.[/quote]

Can someone please explain me this step

$\sum_{cyc}\frac{x^{4}}{xyz(y+z)}\geq\frac{(x^{2}+y^{2}+z^{2})^{2}}{2xyz(x+y+z)}$

Thanks  :wink:

[quote="Leon"]
Can someone please explain me this step

$\sum_{cyc}\frac{x^{4}}{xyz(y+z)}\geq\frac{(x^{2}+y^{2}+z^{2})^{2}}{2xyz(x+y+z)}$

Thanks  :wink:[/quote]

CBS (in Engel form)

See here: http://www.mathlinks.ro/Forum/viewtopic.php?t=15125 for CS in Engel form.


hey guys, i don't know much about tech and computers and stuff so i need some help here..
i want to buy a laptop but i don't know which brand or model..
i will use it just for normal college use, but i am also a musician and i am interested in incorporating some electronic elements in my band and i've seen some bands where the keyboard player has a Mac laptop... what can they do with a this laptop?
do i have to buy an Apple laptop for music purposes or can i buy a regular one that uses Windows?
i cant spend maximum like 1,400 bucks.. it would be cool if you could list some of the best laptops out there (those of you who know about this stuff) ..thanks for your help!!

unless you are used to using a Mac, don't get one. they are good for video editing...not sure about music. If you want a good screen, go fujitsu. For cheap price, HP. I have a Toshiba, pretty well rounded out. iTunes works on Windows..so i guess its fine. Windows comps are compatible with a lot more software and stuff. $1400 seems good enough for a pretty decent laptop..just go visit some stores..check out the laptops and see what you want. off topic..there was this really nice AlienWare laptop (for gaming) that was like$3000 bucks..huge screen, fast processor, lots of RAM...battery life..one hour. plus adapter is brick-sized


In triangle $ABC, AB=\sqrt3, BC=\sqrt4,$ and $CA=\sqrt5$ consider the set of points $P$ in the triangle's interior such that $PA, PB,$ and $PC$ are sides of a triangle.  What is the minimum perimeter of all such triangles?

here's a hint
[hide]
Fermat
[/hide]
and a bigger hint
[hide]
Fermat point
[/hide]


limit
real analysis
real analysis unsolved

$0 < b_0 \leq a_0$.  $a_{n+1} = \frac{a_n + b_n}{2},  b_{n+1} = \sqrt{a_nb_n}$.  It is easy to see that $x := \lim_{n \rightarrow \infty}{a_n} = \lim_{n \rightarrow \infty}{b_n}$ exists.  Find a easy expression of $x$.

I wonder if there is such an easy expression:

[url]http://planetmath.org/encyclopedia/ArithmeticGeometricMean.html[/url]

[quote]The AGM can be used to numerically evaluate elliptic integrals of the first and second kinds. [/quote]I suppose the answer is no. :(


number theory
least common multiple
greatest common divisor

$ \text{lcm}(m,n)+\gcd(m,n)=m+n$, $ m,n\in\mathbb{Z}^{+}$. Prove that either $ m|n$ or $ n|m$.

See SplashD's proof below.  It is nice....

Let $ a=gcd, b=lcm$. We know $ ab=mn$ and we're given $ a+b=m+n$. So $ a+b=m+\frac{ab}{m}\Leftrightarrow (m-a)(m-b)=0$ WLOG $ m=a$, then $ n=b$ and the conclusion follows.

[quote="baz"]Recall if $ m=\prod_{i=1}^{k}p_{i}^{a_{i}}$ and $ n=\prod_{i=1}^{k}p_{i}^{b_{i}}$ where $ p_{i}$ are the distinct prime divisors of $ m$ and $ n$ and the $ a_{i},b_{i}\ge 0$ for all $ i=1,2,\ldots,k$.  Then,
\[ \mbox{lcm}(m,n)=\prod_{i=1}^{k}p_{i}^{\max\{a_{i},b_{i}\}}\mbox{~~and~~}\mbox{gcd}(m,n)=\prod_{i=1}^{k}p_{i}^{\min\{a_{i},b_{i}\}}. \]
For the proof:  Suppose not, i.e. suppose that there exists a prime $ p$ dividing (WLOG) $ m$ and also $ p$ does not divide $ n$.  Then, $ p\mid\mbox{lcm}(m,n)$ and $ p$ does not divide $ \gcd(m,n)$.  Altogether,
\[ p\mid \big( \mbox{lcm}(m,n)-m \big)=n-\gcd(m,n) \]
which results in a contradiction.[/quote]

I don't see why that is a problem. if $ p\not |n$ and $ p\not | \gcd(m,n)$, then why is it a problem if $ p|(n-\gcd(m,n))$? What if $ p=5=m$, $ n=6$?

Furthermore, what if like $ 4|m$, $ 2|n$ and $ 9|n$ and $ 3|m$. You didn't account for powers of primes.

[quote="Altheman"]I don't see why that is a problem. if $ p\not |n$ and $ p\not | \gcd(m,n)$, then why is it a problem if $ p|(n-\gcd(m,n))$? What if $ p=5=m$, $ n=6?$

Furthermore, what if like $ 4|m$, $ 2|n$ and $ 9|n$ and $ 3|m$. You didn't account for powers of primes.[/quote]

 :oops: you are right  :oops:   I don't know what I was thinking.  I was careless, sorry about that.  Thanks for pointing this out.  I will try to edit my post.

[quote="13375P34K43V312"]$ \text{lcm}(m,n)+\gcd(m,n) = m+n$, $ m,n\in\mathbb{Z}^{+}$. Prove that either $ m|n$ or $ n|m$.[/quote]
Wlog, let $ m\le n$. Then $ \mbox{lcm}(m,n)+\gcd(m,n)\ge 2n+1 > m+n$, if $ m\nmid n$. So $ m\mid n$, and then effectively $ \mbox{lcm}(m,n)+\gcd(m,n) = n+m$.