logarithms
superior algebra
superior algebra unsolved

There was this problem from the Romanian TSTs:

[quote="RomTST"]Solve the equation $3^x=2^xy+1$ in positive integers.[/quote]

How do you solve that using rings? I'm dying to know that...

I hope that this question isn't that strange.

i tried not using rings and i got
$x=\log_{\frac{3}{2}} 2y$
is this correct ?

You must solve the equation in [color=blue]positive numbers[/color].

I heard that this problem could be solved with rings, but I don't know how to do it.


function
linear algebra
matrix
algebra
system of equations

Find all functions $ \mathbb{R}\setminus\{0,1\}\to\mathbb{R}$ with $ f(x)\plus{}f\left(1\minus{}\frac 1x\right)\equal{}1\plus{}x.$

[hide="Solution"] The technique is standard.  Let $ g(x) \equal{} 1 \minus{} \frac{1}{x} \equal{} \frac{x \minus{} 1}{x}$.  Verify that

$ g(g(x)) \equal{} \frac{(x \minus{} 1) \minus{} x}{x \minus{} 1} \equal{} \frac{1}{1 \minus{} x}$
$ g(g(g(x))) \equal{} \frac{x}{x \minus{} (x \minus{} 1)} \equal{} x$

We are given $ f(x) \plus{} f(g(x)) \equal{} 1 \plus{} x$.  Repeated transformation $ x \mapsto g(x)$ gives the system of equations

$ f(g(x)) \plus{} f(g(g(x)) \equal{} 1 \plus{} g(x)$
$ f(g(g(x)) \plus{} f(x) \equal{} 1 \plus{} g(g(x))$

Subtracting $ (2)$ from $ (1) \plus{} (3)$ produces

$ 2f(x) \equal{} 1 \plus{} x \plus{} g(g(x)) \minus{} g(x) \equal{} \boxed{ 1 \plus{} x \plus{} \frac{1}{1 \minus{} x} \minus{} \frac{x \minus{} 1}{x} }$ [/hide]

[quote="t0rajir0u"][hide="Solution"] The technique is standard.  Let $ g(x) \equal{} 1 \minus{} \frac {1}{x} \equal{} \frac {x \minus{} 1}{x}$.  Verify that

$ g(g(x)) \equal{} \frac {(x \minus{} 1) \minus{} x}{x \minus{} 1} \equal{} \frac {1}{1 \minus{} x}$
$ g(g(g(x))) \equal{} \frac {x}{x \minus{} (x \minus{} 1)} \equal{} x$

We are given $ f(x) \plus{} f(g(x)) \equal{} 1 \plus{} x$.  Repeated transformation $ x \mapsto g(x)$ gives the system of equations

$ f(g(x)) \plus{} f(g(g(x)) \equal{} 1 \plus{} g(x)$
$ f(g(g(x)) \plus{} f(x) \equal{} 1 \plus{} g(g(x))$

Subtracting $ (2)$ from $ (1) \plus{} (3)$ produces

$ 2f(x) \equal{} 1 \plus{} x \plus{} g(g(x)) \minus{} g(x) \equal{} \boxed{ 1 \plus{} x \plus{} \frac {1}{1 \minus{} x} \minus{} \frac {x \minus{} 1}{x} }$ [/hide][/quote]

I got that same solution (and I wonder if there is another one?).
This problem was posed on the german IMO team qualifications... let's hope the problems are that easy this year :D

[quote="bertram"]I got that same solution (and I wonder if there is another one?).
[/quote]

I have not thought about another solution, but I think that by seeing an expression like $ \frac{1}{x}$, the use of cyclic functions really jumps out at most students.

The motivation behind the solution is knowledge about [url=http://en.wikipedia.org/wiki/M%C3%B6bius_transformation]fractional linear transformations[/url].  See in particular a discussion of the [url=http://en.wikipedia.org/wiki/M%C3%B6bius_transformation#Projective_matrix_representations]matrix representation[/url] of such transformations.  The fact that $ g(g(g(x))) \equal{} x$ corresponds to the fact that

$ \left[ \begin{array}{cc} 1 & \minus{}1 \\ 1 & 0 \end{array} \right]^3 \equal{} \left[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right]$

This is the identifying property of $ g(x)$, so I doubt that another elementary solution exists.

What about:

-1/2, -3/(4t)
t, -1/2

-t-1,-(t^2+t+1)/t
t,t

Is there some extra property that I don't know about?

EDIT: wait ---- does that matrix operation do the composition of that function???

wow thats cool

i remember playing around with these functions on my own a while ago, i didn't know that there was so much cool stuff associated with them !!!

[quote="Altheman"]EDIT: wait ---- does that matrix operation do the composition of that function???[/quote]

Yes, up to multiplication by a constant.  See the second link in my other post.


There is a trapezoid, $ABCD$, such that $AD$ is parallel to $BC, BC=CD=12, \angle{D}=90, AE=10, BC>AD$, and $\angle{ABE}=45$.  What does $CE$ equal?

[quote="Altheman"]the length: CE is not uniquely defined...i have verified by a construction program

maybe there is another condition?[/quote]
I was wonderin' the same thing.

I drew it, and I had this..

[img]http://img295.imageshack.us/img295/1885/traprz0.png[/img]

If I'm correct, that could be the situation, where $\overline{CE}=6$


Given an acute triangle $ABC$ with $O$ as its circumcenter.  Line $AO$ intersects $BC$ at $D$.  Points $E$, $F$ are on $AB$, $AC$ respectively such that $A$, $E$, $D$, $F$ are concyclic. Prove that the length of the projection of line segment $EF$ on side $BC$ does not depend on the positions of $E$ and $F$.

Let $ X$ and $ Y$ be the projections of $ D$ in $ AB$ and $ AC$. Since $ AEDF$ is cyclic, $ \angle DEA \equal{} \angle DFC$, i.e., $ \angle DEX \equal{} \angle DFY$, which means that $ DEX$ and $ DFY$ are similiar. If we proved that the projections of $ EX$ and $ FY$ have the same size, we would know that the projections of $ XY$ and $ EF$ have the same size, which suffices. Now let $ M \equal{} \frac {A \plus{} B}{2}$ and $ N \equal{} \frac {A \plus{} C}{2}$. Also, let $ R \equal{} BO \equal{} CO$. We have

$ \frac {EX}{FY} \equal{} \frac {DX}{DY} \equal{} \frac {OM}{ON} \equal{} \frac {Rcos\angle BCA}{Rcos\angle ABC}$

Hence $ EXcos\angle BAC \equal{} FYcos\angle BCA$, and these are the measures of the desired projections.

Obs.: $ D$ could be an arbitrary point in the line $ AO$.

Let's take the attached picture as reference. First notice that $\angle{FED}=\angle{DAF}=\angle{BEX}$. Now we have

\[XY=EF \sin{\angle{FEX}} = EF \sin(\angle{FED}+\angle{DEX})\]

\[ = EF \left(\frac{BX \cdot EX + EX \cdot DX}{BE \cdot DE}\right)=DB\cos{\angle{BEX}} \cdot \frac{EF}{ED}\]

\[= DB\cos{\angle{BEX}} \cdot \frac{\sin(180-\angle A)}{\sin \angle DAB}\].

Finally, all the terms of the $RHS$ are constant, so $LHS$ must as well be constant.

[I]After a hint to consider $T$:[/i]
Define $\angle EAD = \angle EFD = a$ and $\angle FED = \angle DAF = b$. Let $T$ be the intersection of $(AEF)$ and $BC$. We know that $\angle ABO = a$ and $\angle BOC = 2a + 2b \Longleftrightarrow \angle OBD = 90 - a - b$. Therefore, $\angle ABD = 90 -b $ and $\angle ACD = 90 - a$. Noting that $\sin (90 + b - a) = \sin (90 + a -b)$, $\frac{BD}{BA}\cdot \sin (b) = \frac{DC}{CA} \cdot \sin (a)$
Note that $\frac{BD}{BA} = \frac{\sin (a)}{\sin (90 + b -a)}$ and $\frac{DC}{CA} = \frac{\sin (b)}{\sin (90 + a - b)}$.
$$XB + YC = BE  \sin (b) + CF \sin (a)$$
Note that by Power of a Point, $BE \cdot BA = BT \cdot BD$, so $BE = \frac{BT \cdot BD}{BA}$. By symmetry, $\frac{CT \cdot CD}{AC} = AF$. Thus,
$$XB + YC = \frac{BT \cdot BD}{BA} \sin (b) + \frac{CT \cdot CD}{AC} \sin (a) = \frac{1}{\sin (a) \sin (b)} \left(\frac{BT \cdot BD \sin (b)}{BA} + \frac{CT \cdot CD \sin (a)}{AC} \right)$$
Using what we derived at the beginning,
$$XB + YC = \frac{1}{\sin (a) \sin (b)} \cdot \frac{\sin (a) \sin (b) }{\sin (90 + b - a)} \left(BD + CD \right) = \frac{BC}{\sin (90 + b - a)}$$
It follows that $XB + YC$ is independent of $E$ and $F$, so $BC - (XB + YC) = XY$ is also independent of $E,F$.

[hide=My Solution]Let the length of the projection for some $E$ and $F$ be $l_{E,F}$.
Let $X$ and $Y$ be the projections of $E$ and $F$ on $BC$ respectively.
Consider point $E'$ on $AC$ other than $E$ and $F'$ on $AB$ other than $F$ such that they satisfy the given properties as well.
Since $\square AEDF$ and $\square AE'DF'$ are cyclic quadrilaterals, we have $\angle DEF=\angle DAF=\angle DAF'=\angle DE'F'=\angle DAC=90^\circ-B$. Similarly,  $\angle DFE=\angle DF'E'=90^\circ -C$. Let $K$ is the intersection of $EF$ and $E'F'$. Since $\angle DEF=\angle DEK=\angle DE'F'=\angle DE'K$, we get $\square EE'DK$ as cyclic. Similarly, $\square FF'DK$ is also cyclic.
Let $\angle EDE'=x\Rightarrow\angle EKE'=x$ (as $\square EE'DK$ is cyclic), $\Rightarrow\angle F'KF=x=F'DF$ (as $\square FF'DK$ is cyclic).
Hence, $\angle EDE'=\angle F'DF$.
Let $R_1$ be the circumradius $\square EE'DK$ and $R_2$ be the circumradius of $\square FF'DK$.
Applying Sine Rule,
(i) (in $\circ EE'DK$): $\frac{KD}{\sin (\angle KED)}=\frac{KD}{\sin(90^\circ-B)}=\frac{KD}{\cos B}=2R_1=\frac{EE'}{\sin x}$,
(ii) (in $\circ FF'DK$): $\frac{KD}{\cos C}=\frac{FF'}{\sin x}$.
Dividing both these equations gives $\frac{\cos C}{\cos B}=\frac{EE'}{FF'}\Rightarrow EE'\cos B=FF'\cos C$.
Now,
$EB\cos B=BX$ and $E'B\cos B=BX'$. Subtracting, we get $EE'\cos B=XX'$.
Similarly, we get $FF'\cos C=YY'$.
Hence, $XX'=YY'$. This means that $l_{E',F'}=l_{E,F}+(XX'-YY')=l_{E,F}$.
So, the projection length $l_{EF}$ is equal for two arbitrary pairs of points $E,F$, and hence is independent of them.

Hence proved.[/hide]
[hide=Remark][b]$\textbf{Remark.}$[/b] If the problem would have changed from using circumcenter $O$ to incentre $I$, then the result would be $EE'=FF'$. [/hide]

Figure:

Though it is easy to find the trigonometric solution, does there exist a geometric solution to this?

Let $P$ be the foot of the altitude from $A$, and suppose $\overline{AP}$ intersects $\odot(AEF)$ for the second time again at $Q$.  Further, let $X$ and $Y$ be the projections of $E$ and $F$ onto $\overline{BC}$.  The given problem asks us to show $XY$ is constant.

Because $AO$ and $AP$ are isogonal, $EF\parallel QD$.  Thus
\[
XY = \frac{XY}{PD}\cdot PD = \frac{EF}{QD}\cdot PD = \frac{\sin A}{\sin\angle PAD}\cdot PD,
\]
which is indeed constant.  Yay.


MATHCOUNTS
blogs
function
geometry
geometric transformation
rotation
reflection

Because there is currently a lacking of new MC competitions, I guess I'll make one that's relatively similar. Registration will begin today, and end on August 9. Round 1 will then begin August 12, and will be posted by someone other than me, if possible. This is because I'll be on vacation, and if I wait to post it, you'll have to wait until September. Round 1 will consist of 10 problems, each requiring solutions. Some may only require saying trial and error, but that will still be needed. I'll have 6 rounds, to be explained later. pm me registration, with your name, grade you're going into, and availability from now until around September 20. I hope this goes well!

I don't need real names, and s/n will be there already, so disregard that.

The format of the contest will be as follows:Round 1 will contain all participants, with 10 questions that require solutions. This is the identical format of Round 2. Each question is worth 10 points in these 2 rounds, 3 for correct answer, 7 for solution. The top 12 make it to the playoffs, with top 4 getting a bye the first round. Then, the rest of the playoffs will have 8 people, and a standard single-elimination format. The playoffs will not require solutions, and ties will all be won by whomever sent in answers first. If there are 2^n people answering in a particular playoff round, there will be 9-2n questions. Come on, let's start registering!

[quote="perfect628"]This post is to move the thread to the top, and raise its awareness level. PLEASE SIGN UP!![/quote]

BUMP:

bring up my post

I'm definitely in. Should be interesting.

i am in
here is my info:

akalra1
going into seventh grade
availabilty:[list]weekdays from about 4:00 - 7:00 / 8:30 - 9:30 EST
weekends vary everytime, so i will pm u every week[/list]this week i wont be here on sunday

btw, my school starts August 31 so i will be more busy after that

By the way, I'll be posting and grading the first round because perfect is going to be away.

Just giving people a heads up.

[quote="lotrgreengrapes7926"]Are we still eligible if we're going into high school?  :maybe: :D[/quote]

sure, why not.

[quote="jclarkemathL314159"]No I mean for round 2[/quote]

that's not my problem. I don't even think any were graded. at least i didn't grade any because i don't have the solutions. go ask perfect628

No one gets to know any scores until 9/11, and for 9., remember-some nubers have more than one spelling.

I'm afraid that this comp has not gotten enough responses, so its dead. I'll make a new one some other time.

you should pm a mod to you know lock the topic once the competition is over...its what i do :D

noooo don't kill this competition its awesome  :o

i know..someone shoudl perfect628 that he should continue

can i do round 1 and round 2 to join..i realy want to do it now

[quote="perfect628"]I would, if I could get at least 12 participants...if you want it to continue, I guess I would...if I had some advertisement.[/quote]

Erm, I do not suggest using threads to advertise.  Either put it in your sig, or mass pm people (but not like, random people, just people you know visit the mc forum.

Count me in...I'll go into the playoffs :lol:  :D

that's only 3 people

[quote="perfect628"]I would, if I could get at least 12 participants...if you want it to continue, I guess I would...if I had some advertisement.[/quote]

I sent in my answers for round 1 and will send them in for round 2 if this competition keeps going.

The comp is now dead. However, my new one will have similar questions, so join the amazing race!!!


integration
function
calculus
real analysis
real analysis unsolved

If $b_n = \int_0^1 f^n(x) dx = \int_0^1 |f(x)|^n dx$ then $b_n = \frac{n-1}{n}b_{n-2}$ [*]
Iterating for odd and even indices one can get $\frac{b_{2n+1}}{b_{2n}} = \frac{1}{2n+1} \cdot (\frac{2\cdot4\cdot...\cdot2n}{1\cdot3\cdot...\cdot(2n-1)})^2 \cdot b_1 \to  \frac{\pi}{2}b_1$ [classical result]
Also $\frac{b_{n+1}}{b_n} \to  M = sup f(x)$ [classical result] and from here $b_1 = \frac{2}{\pi}M$

Obs:
The [*] condition is verified also by the classical sequence $a_n = \int_0^{\pi/2} sin^n x dx$ whom behaviour is well known.
From here, a function that verifies conditions a) and b) is $f(x) = sin(\frac{\pi}{2} \cdot x)$ for which M = 1.
I wonder if exists a function for which M < 1.
Can anyone give such an example ?
Mr. Lupas do you have such functions in the pocket ? :)

[quote="mareleG"] ...Can anyone give such an example ?  Mr. Lupas do you have such functions in the pocket ? [/quote]

Please see:
[b]
[1] [/b] A.L., [i]Problem 37, Gazeta Matematica (Perfectionare medodica si metodologica in matematica si informatica=:Seria A),[/i]
 Anul III,nr.3-4,(1982) 86--87.
[b]
[2] [/b] A.L., [i]Solutie la problema 37,Gazeta Matematica  (Perfectionare medodica si metodologica in matematica si informatica=:Seria A),[/i]
 Anul VII,nr.1,(1986) 32--34.
[b]
[3][/b] Andrei Vernescu, [i] Some aspects related to the discrete analogue of differential and integral equations ,[/i]
  Buletinul Stiintific al Universitatii "Politehnica" din Timisoara , Tom 49(63),1,(2004/2005 ?) 20--25.

[quote="mareleG"]Iterating for odd and even indices one can get $\frac{b_{2n+1}}{b_{2n}} = \frac{1}{2n+1} \cdot (\frac{2\cdot4\cdot...\cdot2n}{1\cdot3\cdot...\cdot(2n-1)})^2 \cdot b_1 \to \frac{\pi}{2}b_1$ [classical result][/quote]

How do we prove that classical result?

Its the Wallis formula and here is a proof


analytic geometry
graphing lines
slope
calculus
algebra
polynomial
function

Hello all,
I'm brand new here, and i had a few questions.
I had my freshman year in high school in France.  My school recommended the TI-86 (and as you all know when a school recommends a calculator, all the teachers use that one and know how to use it)  However, i moved right after my freshman year to chicago, and my new school recommends the TI-83/TI-83+.  I was just wondering if there were any benefits to the 86?  Any at all?  It seems to  graph slower, and it seems like simpler tasks are easier on the 83's.  Any help would be great!  I'm currently a junior taking precalculus, and there are lots of graphs involved, and as such it is getting somewhat aggravating due to the slow graphing.

Chris.

I have the TI-86 (I have had it and used it for about 7 years). I prefer it to the TI-83/84+ because of the built in features which are designed for College Math courses such as Linear Algebra, Differential Equations, and Statistics.

All of the features of the TI-86, however, can be accomplished on the TI-84+ with application downloads. And you are right that it is slower, because the processor has not been updated (as far as I know) for many years.

If speed is what you value, and you either don't need or don't mind the application downloads for the slightly more advanced features, then you might consider trading in for the TI-84+. Most American teachers will also be more familiar with the TI-84+.

Thank you so much for your response.

I have one more question.  On the TI-86, there are 4 modes, the Fun, Pol, Param, and DifEq modes.  I have used all of them except for the DifEq mode (perhaps ill use it next year in Calc?)  Anyways, the 83/83+ do not have the DifEq mode, but rather something else.  I take this as a sign that one would not be able to do differential Equations (in college i presume) on the 83/83+ ?

Thank you

Chris.

DiffEq is for graphing slope fields, I think... it's pretty cool, but I haven't used it much during calculus so far this year.

By the way, my favorite thing about the 86 is that it keeps the menus open at the bottom. Saves a lot of keystrokes when you want to repeatedly use something.. The 86 the only calculator I have that keeps the menus open.

Yah, i really like that feature as well.  I just get REALLY REALLY annoyed when im in a test and i have to graph something, and since it graphs so slow i usually hit graph, and then move onto the next problem to wait hehe.
Also, if i plan to go into physics in college, would an 89 be beneficial to me? would i be able to use it at all?

Thanks

Chris.

An 89 is invaluable in calculus. (at least it is for me... very useful for checking answers)
So if you plan on taking calculus-based physics in college (as I presume) it's really going to help.
Plus, it graphs 3D... (yay!) and a lot of other stuff.

Cheers!

I knew that it would help me, i was just wondering if i would be allowed to use it.  I doubt i can use it in my calculus class next year, but in college? i've heard horror stories where you can't use it.

Chris.

It'll depend on the particular teachers you get.  My intro mechanics class last year had no calculator use on exams, for example.

But you'll always be able to use it for homework, so i guess that counts for something.

And checking answers in homework, that is probably valuable as well.

Chris.

The TI-89 has very powerful symbolic algebra capabilities (for a handheld calculator) and for that reason is banned on one kind of college entrance test (the ACT) and in many college courses. For that reason, I so far haven't bought an 89. But the 89 is a wonderful calculator and is recommended for students in my son's math program who didn't already have another calculator. (He has a 83+.)

TI has been pretty smart about confusing the market just enough to sell more than one calculator per student.  ;)  Use what your school recommends, or use what you have if it does most of what the machine your school recommends does and it isn't banned.

Buy him one once he's comfortable (computationally speaking) with calculus.  You can even buy him one a bit earlier than that , but it can breed dependence.  I love my TI-89 - obviously it can't do the stuff Maple can do, but it's fast, compact, and makes no noise - which is why I don't like doing math at computers.

I would get the TI-89 if I were you... it's literally 10 times better than Ti-83+ or Ti-84 or Ti-86, in my opinion. If you want graph fast, Ti-89 graphs about three times as fast as Ti-83+. Plus, it does symbolic algebra, so you can solve EXACTLY most equations that are solvable, factor and expand polynomials, differentiate and integrate (almost) all functions that are diff'ble and integrable. It's the only TI graphing calculator that does symbolic algebra. And it's very easy to use, with those onscreen menus and stuff. And it has "pretty print", so instead of all those parentheses and those "^" and "*" symbols, you see exactly how you would write it with a pencil. Plus it has a much higher screen resolution. Pretty amazing calculator...


inequalities
function
geometry
geometric transformation
conics
parabola
rearrangement inequality

[quote="hien"]Let $a,b,c>1$. The following result true or false:
$\frac{a^2}{b-1}+\frac{b^2}{c-1}+\frac{c^2}{a-1}\geq16$[/quote]
If you are mean to $\frac{a^2}{b-1}+\frac{b^2}{c-1}+\frac{c^2}{a-1}\geq12,$ this is true.

[hide]Regardless of the ordering of the variables, the sequences $(a,b,c) \left(\frac{1}{a-1}, \frac{1}{b-1}, \frac{1}{c-1}\right)$ will always be sorted in opposite directions, so by the rearrangement inequality we have
$\frac{a^2}{b-1} + \frac{b^2}{c-1} + \frac{c^2}{a-1} \geq \frac{a^2}{a-1} + \frac{b^2}{b-1} + \frac{c^2}{c-1}$
The funciton $f(x)= \frac{x^2}{x-1}$ is a translation of $g(x) = \frac{(x+1)^2}{x} = x + 2 + \frac{1}{x}$ by one unit to the right. $g''(x) = \frac{2}{x^3} \geq 0$ for $x\geq 0$ so it is convex on the interval $(0, \infty]$, and $f(x)$ is convex on $(1, \infty]$ .
By Jensen's inequality we have
$\frac{a^2}{a-1} + \frac{b^2}{b-1} + \frac{c^2}{c-1} \geq 3\left(\displaystyle\frac{\displaystyle\frac{(a+b+c)^2}{9}}{\displaystyle\frac{a+b+c}{3}-1}\right) = \frac{(a+b+c)^2}{a+b+c-3}$
The parabola $x^2 - 12x + 36 = h(x)$ has a positive leading coefficient and vertex at $(6,0)$, so it is always $\geq 0$ If we let $a+b+c=x$ then we have
\[\begin{eqnarray*} (a+b+c)^2-12(a+b+c)+36 &\geq& 0\\ (a+b+c)^2 &\geq& 12(a+b+c-3)\\ \frac{(a+b+c)^2}{a+b+c-3} &\geq& 12 \end{eqnarray}\]
Because $a+b+c > 3$

Let I put a hand.
$\frac{a^2}{b-1} + \frac{b^2}{c-1} + \frac{c^2}{a-1} \geq \frac{a^2}{a-1} + \frac{b^2}{b-1} + \frac{c^2}{c-1}$
I will prove $\frac{a^2}{a-1}\geq 4$ equavalent to $(a-2)^2\geq0$
Thanks Arqady for edit my problem

Or we can apply the well-known inequality $\frac{1}{x}+\frac{1}{y}\geq\frac{4}{x+y}$ and then Cauchy. :)


inequalities
geometry
circumcircle
inradius
trigonometry
triangle inequality
trig identities

Prove that in any triangle we have the following inequality:

$\displaystyle 8R^{2}+4Rr-4r^{2} \geq \frac{ (a\sqrt{a}+b\sqrt{b}+c\sqrt{c})^{2}}{a+b+c}.$ (standard notations)

cheers! :D :D

Regarding standard notations I recommend reading [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=21455]these hints.[/url]  :)

Nobody has solved this one yet. Okay, that doesn't mean too much... I rewrite the problem with the definitons of the notations used:

[color=blue][b]Problem.[/b] If ABC is a triangle with circumradius R and inradius r, prove the inequality

$8R^2+4Rr-4r^2\geq\frac{\left(a\sqrt a+b\sqrt b+c\sqrt c\right)^2}{a+b+c}$.[/color]

[i]Solution.[/i] First we note that $8R^2+4Rr-4r^2=4\left(R+r\right)\left(2R-r\right)=4\left(R+r\right)^2\cdot\frac{2R-r}{R+r}$ and $\frac{\left(a\sqrt a+b\sqrt b+c\sqrt c\right)^2}{a+b+c}=\left(bc+ca+ab\right)\frac{\left(a\sqrt a+b\sqrt b+c\sqrt c\right)^2}{\left(a+b+c\right)\left(bc+ca+ab\right)}$, so that the inequality in question becomes

$4\left(R+r\right)^2\cdot\frac{2R-r}{R+r}\geq\left(bc+ca+ab\right)\frac{\left(a\sqrt a+b\sqrt b+c\sqrt c\right)^2}{\left(a+b+c\right)\left(bc+ca+ab\right)}$.

But we have the well-known inequality

$4\left(R+r\right)^2\geq bc+ca+ab$;

a proof of this inequality will be given below. Thus, it remains to show the following inequality:

$\frac{2R-r}{R+r}\geq\frac{\left(a\sqrt a+b\sqrt b+c\sqrt c\right)^2}{\left(a+b+c\right)\left(bc+ca+ab\right)}$.

By the Cauchy-Schwarz inequality,

$\left(a^2+b^2+c^2\right)\left(\left(\sqrt a\right)^2+\left(\sqrt b\right)^2+\left(\sqrt c\right)^2\right)\geq\left(a\sqrt a+b\sqrt b+c\sqrt c\right)^2$;

in other words,

$\left(a^2+b^2+c^2\right)\left(a+b+c\right)\geq\left(a\sqrt a+b\sqrt b+c\sqrt c\right)^2$,

so that

$a^2+b^2+c^2\geq\frac{\left(a\sqrt a+b\sqrt b+c\sqrt c\right)^2}{a+b+c}$

and

$\frac{a^2+b^2+c^2}{bc+ca+ab}\geq\frac{\left(a\sqrt a+b\sqrt b+c\sqrt c\right)^2}{\left(a+b+c\right)\left(bc+ca+ab\right)}$.

Hence, instead of proving the inequality

$\frac{2R-r}{R+r}\geq\frac{\left(a\sqrt a+b\sqrt b+c\sqrt c\right)^2}{\left(a+b+c\right)\left(bc+ca+ab\right)}$,

it is enough to verify the sharper inequality

$\frac{2R-r}{R+r}\geq\frac{a^2+b^2+c^2}{bc+ca+ab}$.

In order to prove this inequality, we denote x = s - a, y = s - b and z = s - c, where $s=\frac{a+b+c}{2}$ is the semiperimeter of triangle ABC. Then, y + z = (s - b) + (s - c) = 2s - (b + c) = (a + b + c) - (b + c) = a and similarly z + x = b and x + y = c. Thus, x + y + z = x + (y + z) = (s - a) + a = s. Also, by the triangle inequality, b + c > a, so that b + c - a > 0, and thus $x=s-a=\frac{a+b+c}{2}-a=\frac{b+c-a}{2}>0$ and similarly y > 0 and z > 0. Thus, the numbers x, y, z are all positive.

If $\Delta$ is the area of triangle ABC, then, by a well-known formula, the circumradius R of triangle ABC equals $R=\frac{abc}{4\Delta}$. On the other hand, by another well-known formula, $\Delta=sr$. Hence, $R=\frac{abc}{4sr}$. Therefore, $\frac{R}{r}=\frac{abc}{4sr^2}$. Thus,

$\frac{2R-r}{R+r}=\frac{2\frac{R}{r}-1}{\frac{R}{r}+1}=\frac{2\frac{abc}{4sr^2}-1}{\frac{abc}{4sr^2}+1}$.

Using this formula and the equations a = y + z, b = z + x, c = x + y and s = x + y + z obtained above, we can find after some boring calculations

$\frac{2R-r}{R+r}-\frac{a^2+b^2+c^2}{bc+ca+ab}=\frac{2\left(x+y+z\right)\left(\left(zx-xy\right)^2+\left(xy-yz\right)^2+\left(yz-zx\right)^2\right)}{\left(yz^2+zy^2+zx^2+xz^2+xy^2+yx^2+6xyz\right)\left(x^2+y^2+z^2+3yz+3zx+3xy\right)}$.

Hence, of course,

$\frac{2R-r}{R+r}-\frac{a^2+b^2+c^2}{bc+ca+ab}\geq 0$,

so that $\frac{2R-r}{R+r}\geq\frac{a^2+b^2+c^2}{bc+ca+ab}$. And this completes the proof of our inequality.

Ah, in the above solution I promised to give a proof for the inequality $4\left(R+r\right)^2\geq bc+ca+ab$. Here is one possible proof:

By the extended law of sines, a = 2R sin A, b = 2R sin B and c = 2R sin C. By a formula from trigonometry, R + r = R (cos A + cos B + cos C). Hence, the inequality $4\left(R+r\right)^2\geq bc+ca+ab$ is equivalent to

$4\left(R\left(\cos A+\cos B+\cos C\right)\right)^2$
$\geq 2R\sin B\cdot 2R\sin C+2R\sin C\cdot 2R\sin A+2R\sin A\cdot 2R\sin B$.

Upon division by $4R^2$, this becomes

$\left(\cos A+\cos B+\cos C\right)^2\geq \sin B\cdot\sin C+\sin C\cdot\sin A+\sin A\cdot\sin B$.

But this inequality was proven in http://www.mathlinks.ro/Forum/viewtopic.php?t=5716 . Hence, the inequality $4\left(R+r\right)^2\geq bc+ca+ab$ is proven.

  Darij

Nice solution Darij.  :lol:
But, isn't it a bit long ??  :?
Anyway your solutions are always long  :D

I am sure $8R^2+4Rr-4r^2$ is something with the sides of the triangle $ab+bc+ca$ or $a^2+b^2+c^2$, Something like that!

From  BCS  inequality  we  have  [tex] 2(s^2-r^2-4Rr)=a^2+b^2+c^2\geq\frac{a\sqrt{a}+b\sqrt{b}+c\sqrt{c}}{a+b+c} [/tex] . So its  enought  to prove  that
[tex]8R^2+4Rr-4r^2\geq2(s^2-r^2-4Rr) [/tex]  or  [tex]4R^2+Rr-r^2\geq s^2[/tex] .But this  comes  from  [tex]4R^2+4Rr+3r^2\geq s^2\Longleftrightarrow IH^2\geq0 [/tex]  Well  [tex]  s [/tex]  is  semiperimeter.


inequalities
geometry
circumcircle
inequalities unsolved

Let $ P$ be a point in the plane of a triangle $ ABC$. Let $ R$ be the circumradius of triangle $ ABC$. Prove that:
$ \frac {PA}{BC^2} \plus{} \frac {PB}{CA^2} \plus{} \frac {PC}{AB^2} \ge \frac {1}{R}$.

I think it was posted before by hxtung. Apparently, it's from the USA IMO training as well :).

Yes, it was posted by Hxtung:
http://www.mathlinks.ro/Forum/viewtopic.php?t=6073 . And I posted a beautiful solution, however not from me...

  Darij


Hello,I'm a new member of http://www.mathlinks.ro.
 This problem isn't easy.
 Denote $ a_n$ the nearest natural number of  $ \sqrt {n}$(where n is a natural number).Find the value of 1/a_1 + 1/a_2 + ... + 1/a_1980

[quote="gauss202"]Compute the first few examples of $ a_n$ and see if you notice a pattern.[/quote]
 For examples:you can see $ \sqrt{1}\equal{}1\Rightarrow a_{1}\equal{}1$
                                  and $ \sqrt{2}\approx 1,414213562\Rightarrow a_{2}\equal{}1$
                                  but $ \sqrt{3}\approx 1,732050808\Rightarrow a_{3}\equal{}2$

[quote="gauss202"]you'll need more examples than that to see the pattern.

[hide="A way to generate the sequence without actually computing square roots:"]

$ \text{Round}(\sqrt {n}) \equal{} k$, $ k$ a positive integer, means that $ k \minus{} \frac {1}{2} \leq \sqrt {n} < k \plus{} \frac {1}{2}$
$ \Leftrightarrow k^2 \minus{} k \plus{} \frac {1}{4} \leq n < k \plus{} k \plus{} \frac {1}{4}$
$ \Leftrightarrow k^2 \minus{} k < n \leq k^2 \plus{} k$
$ \Leftrightarrow k(k \minus{} 1) < n \leq k(k \plus{} 1)$

Now you can count which values of n go to which values of k very quickly.[/hide][/quote]
in fact,this problem was solved by me.Your solution is similar to mine.Do you think this problem is hard?

[quote="lovemathvn"]in fact,this problem was solved by me.Your solution is similar to mine.Do you think this problem is hard?[/quote]Not that hard, but, we would like to see your solution so we can rate its validity. Perhaps there is a AoPSer who has a different solution than yours or suggest another approach that you might learn something new ;)

[quote="lovemathvn"]
in fact,this problem was solved by me.Your solution is similar to mine.Do you think this problem is hard?[/quote]

I actually hate when people refer to problems as easy or hard (even though I've been guilty of it myself), because those are such relative terms. Just because one person finds a problem easy doesn't mean that other people will or should, and just because one person finds a problems hard doesn't mean that other people will or should.

[hide="A small but valuable hint"]$ \frac{1}{\sqrt{k}}\equal{}\frac{2}{\sqrt{k}\plus{}\sqrt{k}} < \frac{2}{\sqrt{k}\plus{}\sqrt{k\minus{}1}}$

Taking the conjugate, we obtain:

$ \frac{1}{\sqrt{k}} < 2(\sqrt{k}\minus{} \sqrt{k\minus{}1})$
It's easy to sum this

And similarly
$ \frac{1}{\sqrt{k}} \equal{} \frac{2}{(\sqrt{k} \plus{} \sqrt{k})} > \frac{2}{ \sqrt{k\plus{}1} \plus{} \sqrt{k}} \equal{} 2(\sqrt{k\plus{}1} \minus{} \sqrt{k})$
Sorry dont have the time to post whole solution
[/hide]

[quote="varunrocks"]As n reaches infinite the value of this is expression is slightly less than 1.674[/quote]Harmonic series is divergent, and so is this one...

[quote="varunrocks"]As n reaches infinite the value of this is expression is slightly less than 1.674[/quote]
my response plans is 88.


Set systems
IMO Shortlist
combinatorics
IMO Longlist

In a group of interpreters each one speaks one of several foreign languages, 24 of them speak Japanese, 24 Malaysian, 24 Farsi. Prove that it is possible to select a subgroup in which exactly 12 interpreters speak Japanese, exactly 12 speak Malaysian and exactly 12 speak Farsi.

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

[quote="orl"][b]remark by Darij Grinberg[/b]: However, it is not always possible to find such a subset from an odd number of translators.[/quote]

This is not a remark by me, but a sentence which stood in brackets in the original book (Morozova / Petrakov). I am not too sure about the meaning of this sentence.

  Darij


function
algorithm
combinatorics unsolved
combinatorics

Let n and k be integers with $ 0 \leq k < \frac{n}{2}$. Initially let A be the sequence of subsets of {1,2,...n} with exactly k elements, and B the sequence of subsets of {1,2,...n} with exactly k+1 elements, both arranged in lexicographic  (dictionary) order. Now let S be the first element in A. If there is a T in B such that S is a subset of T, remove S from A and the first such T from B, and repeat this process as long as A is nonempty; otherwise, stop. Prove that this process terminates with A empty.

[MOP 2007]

I worked on this for quite some hours; I was only able to show the existence of a one-to-one function f from A to B with the property that T is a subset of f(T).

Does this process necessarily terminate? I think according to your wording there's cases that we just stop with both A and B nonempty.

Else, since A always has 1 less element than B, is the conclusion trivial?

While there is always a matching, any particular algorithm may or may not be able to find it without backtracking.  For example, suppose we take $ k \equal{} 2$, $ n \equal{} 5$.   The algorithm matches
\[ 12 \rightarrow 123, 13 \rightarrow 134, 14 \rightarrow 124, 15 \rightarrow 125, 23 \rightarrow 234 24 \rightarrow 245, 25 \rightarrow 235, 34 \rightarrow 345, 35 \rightarrow 135, 45 \rightarrow 145\]
and finishes with $ A$ empty (and in this case $ B$ empty as well, though in general some sets may be left over in $ B$).  However, suppose we changed the algorithm to remove the [b]last[/b] such $ B$ instead of the first.  Then $ 12$ would correspond to $ 125$, $ 13$ would correspond to $ 135$, and $ 14$ would correspond to $ 145$.  Now when we go to match $ 15$, all of the subsets containing $ 1$ and $ 5$ are already taken, so we can't do it.  The point is to show that this problem never happens with the original algorithm.


OK, here's my situation:

I'm pretty slow worker. I like to investigate every single more challenging problem before I work it out, so it take longer. I am thinking of taking AMC/AIME... And I am pretty worried because under tense conditions I think slower and less efficiently. Although I get 8-11's easily on practice AIME's I think that's all going to change when taking the real AIME..Any suggestions for practice or personal advice that would helpful? Sorry I am really worried because I need to make USAMO.. or else I would look down at myself, and beat the **** out of myself. Please any advice is helpful..

- Jimmy Hu

[quote="joml88"][quote="jimhu"] Sorry I am really worried because I need to make USAMO.. or else I would look down at myself, and beat the **** out of myself. Please any advice is helpful..

- Jimmy Hu[/quote]

Whoa, calm down.  The tests should be fun and recreational.  You should be doing math for the love of it, not to make USAMO.  You might want to rethink why you're studying math.  Are you doing it because you love it or are you trying to meet some goals?  To get as far as you already are, you must love math (or so I assume).  So try and take it easy on yourself.[/quote]

yea I like math.. what I am trying to say is that i guess those days spending on practicing aime's will be wasted if i easily go too slow or do something wrong.. And am worried because I don't have good grades.. that's why I need to do something..

The grades are more of an issue, as they can affect college acceptance, etc., but math competitions have nothing to do with it. The whole point is to have fun solving problems. If you consider time spent on practice AIME's to be wasted if you don't qualify for USAMO, then I agree with joml - you should rethink what you're intending to do.

[quote="chess64"]Wait isn't he talking about grades on the AIME (as in scores)? I don't think he was talking about school...[/quote]

In fact I am talking about school.. I have 1 C and 1 B, and they are really hard to raise.. so yea...


Find the equation of the conic tangent to the lines x - 1 = 0; x - 2 = 0; y - 1 = 0; 2x - y = 0; y - 3 = 0:

圆锥曲线是三年以前学的，早就忘了。不过，求轨迹的问题一般可以用几何画板。


Let a hexagon ABCDEF satisfying $ \angle A\plus{}\angle C\plus{}\angle E\equal{}\angle B\plus{}\angle D\plus{}\angle F$ and $ AB.CD.EF\equal{}BC.DE.FA$.
Prove that $ AB.CE.FD\equal{}BF.DE.CA$

Construct Q outside ABCDEF
Q satisfy  △QEF∽△CDE
We obtain AB/BC=FA/FQ  ∠QFA=∠B  =>  △AQF∽△ACB
Furthermore  △QCE∽△FED   △QAC∽△FAB
=> CE/EQ=ED/EF   FB/QC=AB/AC  CE/BF=ED*EQ*AC / AB*QC*EF =ED*AC/FD*AB
CE*FD*AB=BF*ED*AC


I found a (text)book from the 30s or 40s called [i]Advanced Calculus[/i] at Border's and picked it up to use next year (in 12th grade). One of the exercises was this: Does $f(x)=|x|^3\in C^2$? I don't know if the notation is still used, but the book defines $C$ to be the class of continuous functions and gives the definition $f(x)\in C^n \Leftrightarrow f^{(n)}(x)\in C$. So basically the question is whether or not $f(x)$ has a continuous second derivative. I'm pretty sure the answer is yes, but there are a few details I'm not sure about. First, what is $f'(x)$? Based on what I learned in Calculus AB last year it should be $f'(x)=3|x|^2\left(\frac{|x|}{x}\right)=3|x|^2\left(\frac{x}{|x|}\right)=3|x|x, x\ne0$. But this is undefined at $x=0$ it seems. That may not seem like a problem but if you visualize the function $f$ and its first derivative there should be no discontinuities. This led me to ask the following question: Suppose you have $f(x)=|x|-|x| (=0)$. Then $f'(x)=\frac{|x|}{x}-\frac{|x|}{x}$, but this is also defined at 0, even though $f(x)=0$ and $f'(x)=0$ continuously. It's a simple question but I'm very puzzled by it. Thanks in advance.

Using the piecewise definition, we get

$f''(x) = 6x$ for $x\geq 0$ and $f''(x) = -6x$ for $x<0$

It is clear that $\lim_{x\rightarrow 0} f''(x) = 0$, therefore $f''$ is continuous.

As for your second question, $(f+g)'(x)=f'(x)+g'(x)$ only under the assumption that $f,g$ are differentiable at $x$.

[quote="blahblahblah"]Using the piecewise definition, we get

$f''(x) = 6x$ for $x\geq 0$ and $f''(x) = -6x$ for $x<0$

It is clear that $\lim_{x\rightarrow 0} f''(x) = 0$, therefore $f''$ is continuous.[/quote]I realize that much, but what if i wanted to find $f'$ and $f''$ without the piecewise definition? I believe $f'(x)=3|x|x$ and $f''(x)=6|x|$, but my problem is that in calculating them I canceled out $\frac{|x|}{|x|}$ in each one.[quote]As for your second question, $(f+g)'(x)=f'(x)+g'(x)$ only under the assumption that $f,g$ are differentiable at $x$.[/quote]Sounds about right.

The same thing applies when you use the chain rule -  $(fog)'(x) = f'(g(x))g'(x)$ as long as g is differentiable at x and f is differentiable at $g(x)$.. So you get $f'(x) = 3|x|x \hbox{ for } x\ne 0$, and you can calculate $f'(0)$ from first principles.

[quote="TripleM"]...you can calculate $f'(0)$ from first principles.[/quote]Should I know what "first principles" means?

Thats just the way you actually calculate derivatives from scratch. How do you know that when you take the derivative of $x^2$ you get $2x$?
The definition of the the derivative is
$f'(x) = \lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}$

In this case, $f'(0) = \lim_{h\rightarrow 0} \frac{\abs{|h|^3}-0}{h} = 0$.


geometry
incenter
circumcircle
inradius
geometry unsolved

[u]  [i]Problem :[/i][/u]
  [i]Let $ AA' , BB' , CC'$ be the inner angled-bisectors of a triangle $ ABC$
   The incenter , circumcenter , the inradius, circumradius of  which are respectively $ I , O, r , R$
       Let Q be the center of the escribed circle of $ ABC$ in the angle $ A$
 Prove that :
 a/ $ IK \equal{} \frac{rR}{OQ}$ where $ IK$ is the distance form $ I$  to $ B'C'$
 b/  $ IA' \plus{}IB' \plus{} IC' \geq 6r.\sqrt{\frac{ 3R}{ 11R \plus{} 2r}}$[/i]

[i]   Love An Forever[/i]

[i]  It is a very old and nice problem
      Noone can help me? :( [/i]


geometry
rectangle
analytic geometry
graphing lines
slope
circumcircle

Two nonadjacent vertices of a rectangle are $(4,3)$ and $(-4,-3)$, and the coordinates of the other two vertices are integers. The number of such rectangles is

$\textbf{(A)}\ 1 \qquad
\textbf{(B)}\ 2 \qquad
\textbf{(C)}\ 3 \qquad
\textbf{(D)}\ 4 \qquad
\textbf{(E)}\ 5$

[hide]
any point on the circle with radius $\sqrt{4^2+3^2} = 5$ and center at origin will satisfy the conditions because a triangle that has a side as a diameter of the circle and the other point on the circumference of the circle will be a right triangle.
any point on the circle and the point reflected across the origin would make the four points of the rectangle
counting only the points below the line $y=\frac{3}{4}x$,
$(-3,-4) (0,-5) (3,-4) (4,-3) (5,0)$
thus there are five solutions
[/hide]

[hide="Answer"]If one vertex has integer coordinates, the opposite will as well. So, we look at the vertex below $(4,3)$ and right of $(-4,-3)$ and label it $(x,y)$. We have $\frac{3-y}{4-x}=\frac{y-3}{x-4}=-\frac{1}{\frac{y-(-3)}{x-(-4)}}=-\frac{x+4}{y+3}$ because the slopes must be perpendicular. From there, we have $-(x+4)(x-4)=(y+3)(y-3)\Rightarrow 16-x^2=y^2-9\Rightarrow x^2+y^2=25$. This gives integer solutions of $(\pm 3,\pm 4)$, $(\pm 5,0)$, and $(0, \pm 5)$, but this corresponds to only $3$ possible rectangles. $\boxed{C}$[/hide]

[hide]Connect the two points and let the line connecting them be the diameter of a circle.  The origin is the center and the radius is 5, so the equation of the circle is $x^2+y^2=25$.  We can select any point on the circle above the diameter to be the third vertex of the rectangle.  There are two integer points on the circle in each quadrant, as well as two integer points on each axis.  Therefore, there are 12 integer points on the circle.  Two of these points are on the diameter, and half of the remaining 10 are below the diameter, so we have 5 points to choose from when choosing our third vertex.  There are 5 rectangles.[/hide]

[quote="minsoens"]counting only the points below the line $ y = \frac {3}{4}x$,
$ ( - 3, - 4) (0, - 5) (3, - 4) (4, - 3) (5,0)$
thus there are five solutions
[/quote]

I don't get why this is done.  Specifically, how do we find $ ( - 3, - 4) (0, - 5) (3, - 4) (4, - 3) (5,0)$.

Thanks a lot!

The radius of the circumscribed circle is 5, centered at the origin.  By looking for integer coordinates, we are really looking for pythagorean triples.  For a hypotheneuse of 5, I only know of the degenerative $ 0^2\plus{}5^2\equal{}5^2$ and the familiar $ 3^2\plus{}4^2\equal{}5^2$.  The five solutions are deduced by using 0 and 5, and 3 and 4, both in the x and y coordinates and both positive and negative.  The five mentioned are shown in red.

Why are we only counting points under $ y\equal{}\frac{3}{4}$?

[quote="314.math"]Why are we only counting points under $ y \equal{} \frac {3}{4}$?[/quote]

because those points directly determine the point over y=.75 y.

i have answered about 10 of your little questoins in the past day.

[quote="314.math"][quote="jli"]because those points directly determine the point over y=.75 y. [/quote]

I still don't get it.  could you please clarify?  Thanks.

Thank you so much for answering all my questions.[/quote]

because it is a rectangle.  if you have 3 points, the 4th one must be determined.

my point is that you need to spend more time thinking about these.  at least 20-30 minutes before you ask because the questions are relatively simple, no offense.  it might be better if you post in a easier forum, like HSB :wink:

either that or youre trying to get your post count up.

Let the vertex of the rectangle be $ (a,b)$.
Then $ (3 \plus{} b)/(4 \plus{} a) \equal{} (a \minus{} 4)/(3 \minus{} b)\implies a^2 \plus{} b^2 \equal{} 25$
The possible choices are $ (\pm3,\pm4), (\pm4,\pm3), \pm(0,5), \pm(5,0)$, giving a total of $ 2*2*2 \plus{} 2*2 \equal{} 12$ coordinates. But we cannot have $ (4,3)$ and $ ( \minus{} 4, \minus{} 3)$ as that would make the slope 0/0. So the answer is $ (12 \minus{} 2)/2 \equal{} 5$.


modular arithmetic
abstract algebra
number theory
relatively prime
prime numbers
greatest common divisor

Find a positive integer $a$, such that the equation $4x+7y=a$ does not have non-negative-integer solutions; but for any positive integer $b>a$, the equation $4x+7y=b$ has non-negative-integer solutions

maybe he means the largest integer that cannot be expressed as 4a+7b?

...

so anything 0 mod 4 can be
anything 1 mod 4 > 21 can be
anything 2 mod 4 > 10 can be
anything 3 mod 4 > 3 can be

so $a=21$ I think...

what's w/ chicken mcnuggets EH

It says nonnegative integer solutions, not positive integer solutions. So $18, 19, 20, 21$ are all possible.

[hide="Solution for this problem"]
We see that these numbers are expressible as $4x+7y$ for $(x,y) \in \left(\mathbb{N^{*}}\right)^{2}$ :
All numbers $\geq 0$ and $\equiv 0 \bmod 4$.
All numbers $\geq 21$ and $\equiv 1 \bmod 4$.
All numbers $\geq 14$ and $\equiv 2 \bmod 4$.
All numbers $\geq 7$ and $\equiv 3 \bmod 4$.

So, $a$ is most likely $\equiv 1 \bmod 4$. Checking, we immediately find $\boxed{a=17}$ as our solution. This is the equivalent solution given by the Chicken McNugget Theorem.[/hide]

[i]Follow-up question:[/i]

For positive integer constants $a$ and $b$, find a positive integer $m$, such that the equation $ax+by=m$ does not have positive-integer solutions; but for any positive integer $n>m$, the equation $ax+by=n$ has positive-integer solutions.

Consider the equation modulo $4$. It becomes $3y \equiv 0 \pmod{4}$. With $y=0,1,2,3,4$ we get $3y$ congruent to $0,3,2,1,0$ respectively. Thus every integer greater than or equal to $7(3)$ and congruent to $0,3,2,$ or $1$ modulo $4$ can be written in this form, that is, every integer greater than or equal to $21$ that can be written in this form.

However, every integer congruent to $1$ and less than $21$ cannot be written in this form (because $21$ was the first integer of this form to be congruent to $1$ modulo 4). The largest such integer that cannot be written in this form is $21-4=17$. (notice how we can get every integer congruent to $0,2,$or$3$ that is less than 21)

The answer to the follow up question would just be $ab-a-b$, proven in a similar way.

[quote="cosinator"]Maybe Chicken McNugget?[/quote]
[quote="BanishedTraitor"]I don't think so. I  think the original problem is correct and that $a$ is just obtained using the Chicken McNugget Formula.[/quote]
What is the Chicken McNugget Formula?

[quote="paladin8"][quote="cincodemayo5590"][i]Follow-up question:[/i]

For positive integer constants $a$ and $b$, find a positive integer $m$, such that the equation $ax+by=m$ does not have positive-integer solutions; but for any positive integer $n>m$, the equation $ax+by=n$ has positive-integer solutions.[/quote]

This is very difficult when $(a, b) > 1$ :P.[/quote]

 :D   Well, I've got a feeling that someone around here, with all of the math-skills floating around, would be able to find a pair $(x,y)$ that satisfies $2x+4y=11$.

Is there an elementary way (without using mod) to do it?

[quote="davidyko"]If you remember that the largest positive number that cannot be expressed as a sum of some multiple of two relatively prime numbers $a$ and $b$ is $ab-a-b$ as someone said, you should be fine.[/quote]
But how to prove it (without mod)? I have never learned it.

You could do some reasoning about sequences that would be basically equivalent to modular arithmetic.  The result is distinctly number-theoretic.

[quote="davidyko"]If you remember that the largest positive number that cannot be expressed as a sum of some multiple of two relatively prime numbers $ a$ and $ b$ is $ ab \minus{} a \minus{} b$ as someone said, you should be fine.[/quote]
The following is a proof. If there is any mistake, please point me out.

First, we have to prove $ au \plus{} bv\neq ab \minus{} a \minus{} b$ for any $ u,v\in\mathbb N_0$.
Assume that $ \exists u,v\in\mathbb N_0$ such that $ au \plus{} bv \equal{} ab \minus{} a \minus{} b$.
Then for such $ u,v$, we have $ a(b \minus{} 1 \minus{} u) \equal{} b(v \plus{} 1)$.
Since $ a,b\in\mathbb N$ are relatively prime, the greatest common divisor $ \gcd(a,b) \equal{} 1$. Thus $ a|v \plus{} 1$.
But $ a$ and $ v \plus{} 1$ are positive, so $ a\leq v \plus{} 1$. Similarly, we have $ b\leq u \plus{} 1$.
Then we get $ ab \equal{} au \plus{} bv \plus{} a \plus{} b \equal{} a(b \plus{} 1) \plus{} b(v \plus{} 1)\geq ab \plus{} ab \equal{} 2ab$,
which is a contradiction.

Now we prove any integers greater than $ ab \minus{} a \minus{} b$ are expressible as the form $ au \plus{} bv$, where $ u,v\in\mathbb N_0$.
Consider the following $ b$ integers: $ n, n \minus{} a, n \minus{} 2a, \cdots, n \minus{} (b \minus{} 1)a$, where $ n\geq ab \minus{} a \minus{} b \plus{} 1$.
Then the smallest one is $ n \minus{} (b \minus{} 1)a\geq ab \minus{} a \minus{} b \plus{} 1 \minus{} (b \minus{} 1)a \equal{} \minus{} b \plus{} 1$.
Since $ a$ and $ b$ are relatively prime, the above $ b$ integers are not congruent to each other module $ b$.
Hence there is exactly one of them is a multiple of $ b$, denoted by $ n \minus{} ax \equal{} yb$, where $ 0\leq x\leq b \minus{} 1$.
By above, we have $ yb \equal{} n \minus{} ax\geq n \minus{} (b \minus{} 1)a\geq \minus{} b \plus{} 1$,
i.e. $ y\geq \minus{} 1 \plus{} 1/b > \minus{} 1$. So the integer $ y\geq0$.
Thus $ n \equal{} ax \plus{} by$, where $ x,y$ are nonnegative integers.


Or more generally $f(n)=n+a$ ($a$ odd) for $n$ even and $f(n)=n-a+2$ for $n$ odd ;)
But: has $f$ some other property¿ where does $n$ come from¿ (integer/natural assumed above)
And please correct the title...

We have a not hard problems following :

Find all $T \in N$ such that there exits a function $f: N\to N$ satisfied : $f(f(n))=n+T \ \ \forall n \in N$


limit
trigonometry
calculus
calculus computations

Why is the following limit $0$ ?

$\lim_{x\to\infty} \frac{sin(x)}{x} = 0$

Since $-1\leq \sin x\leq 1,$ for $x\neq 0,$ we have $\left|\frac{\sin x}{x}\right|\leq \left|\frac{1}{x}\right|.$

Then you can use Archimedes' axiom,or squeezing theorem.

you can also say that $\sin(x)$ is always bounded by $[-1,1]$ and since the denominator is increasing w/o bound, the limit is $0$

[quote="Kalle"]Why use something like that when it's obvious from the epsilon-delta definition?[/quote]

Because Japanese high school students are not allowed to use $\epsilon-\delta$.

[quote="NaZAf"]Can any of you give a formal demonstration?[/quote]

Since $-1\leq \sin x\leq 1,$ for $x>0\ (\because x\rightarrow \infty),$ we have $-\frac{1}{x}\leq \frac{\sin x}{x}\leq \frac{1}{x}.$ thus by $\lim_{x\rightarrow \infty} \frac{1}{x}=-\lim_{x\rightarrow \infty} \frac{1}{x}=0,$ yielding $\lim_{x\rightarrow \infty} \frac{\sin x}{x}=0.\ Q.E.D.$

[quote="NaZAf"]Thanks kunny!

I see, but the squeeze theorem does not work for $\lim_{x \to 0}\frac{sin(x)}{x}$ , does it?[/quote]

Yes, in this case you can't use this theorem. Sorry I confused to answer tag question. :lol:


In a country there are $ n\ge3$ cities. The highways are designed as follows:
(1) Between any two cities there is one and only one one-way direct highway.
(2) If there is a direct highway from city x to city y, and a direct highway from city y to city z,
then there is a direct highway from city z to city x.
How many cities are there in the country? Justify your answer.

[hide]We must have $ n \equal{} 3$. Suppose $ A$, $ B$, $ C$, and $ D$ are four distinct cities such that A -> B, and B -> C. (where X -> Y means there is a direct highway from X to Y.) Condition 2 tells us C -> A.

By (1), either C->D or D->C. If C->D, then (2) tells us that D->B. By (1), either A->D or D->A. If D->A, then (2) on DAB tells us that B->D, which is a contradiction. If A->D, then (2) on DAB tells us that B->A, which is a contradiction.

If D->C, then (2) tells us that A->D. By (1), B->D or D->B. If B->D, then (2) on BDC yields C->B, a contradiction. If D->B, then (2) on DBC yields C->D, again a contradiction. Hence, we cannot have four cities.

Clearly, A -> B -> C -> A satisfies the condition for $ n \equal{} 3$, resulting in a final answer of $ n \equal{} 3$.

Sorry about the poorly written proof. [/hide]


AMC
AIME
geometry
rectangle
probability
analytic geometry
graphing lines

A circle of radius 1 is randomly placed in a 15-by-36 rectangle $ABCD$ so that the circle lies completely within the rectangle. Given that the probability that the circle will not touch diagonal $AC$ is $m/n$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.

[hide="probably wrong"]
Clearly the boundaries are going to be two triangles inside the rectangle, each 1 unit away from the diagonal and sides.  To find the vertices, we can draw circles of radius 1, tangent to the sides and diagonal.  Let circle x touch AD and AC, and draw a line parallel to AB tangent to x, making a right triangle with incircle x.  Similarly, let circle y touch DC and AC and tangent line parallel to AC make right triangle with incircle y (x and y have radius 1).  Both right triangles are similar to the 15x36x39 triangle we have.  Using the formula for incircles in right triangles, we know
$5k + 12k - 13k = 2$, where $k$ is the scale factor of the smaller right trianles to a normal 5-12-13 triangle.  Solving, $k= \frac 1 2$, so the legs of the triangles are $\frac 5 2$ and $6$.
So the legs of the boundary triangle (the one with sides each 1 away from the diagonal and sides) are going to be $15 - (\frac 5 2 - 1) - 1$ and $36 - (6 - 1) - 1$, or $\frac{25}2$ and $30$.  Note that I subtracted the radius from the side length of the triangle, and also subtracted 1 because it must be 1 away from the edge.  So the area of the triangle is $\frac{375}2$, and there are two, so the total area is $375$.  To find the probability, we divide by the total possible area.  Since we are given the circle is entirely inside the rectangle, the possible area is $13 \cdot 34 = 442$, so the final answer is $\frac{375}{442}$ which isn't very nice to look at, but is perfectly reasonable for an AIME problem.
[/hide]

That's correct, but how are you supposed to bubble fractions in AIME? I always thought it was only integers? :?

On the AIME it will say "The probability can be expressed in the form $\frac mn$.  Find $m+n$."  On Kalva, they reword the questions...thus the answers wont all be between 0 and 999, integers only.

[hide="Solution"]
[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=658[/img]

Let the bisector of $\angle CAD$ be $AE$, with $E$ on $CD$. By the angle bisector theorem, $DE=36/5$. Since $\triangle AOR \sim \triangle AED$ ($O$ is the center of the circle), we find that $AR=5$ since $OR=1$. Also $AT=35$ so $RT=OQ=30$.

We can apply the same principle again to find that $PT=27/2$, and since $QT=1$, we find that $PQ=27/2-1=25/2$. The locus of all possible centers of the circle on this "half" of the rectangle is triangle $\triangle OPQ$. There exists another congruent triangle that is symmetric over $AC$ that has the same area as triangle $\triangle OPQ$. $\triangle APQ$ has area $\frac{1}{2}\cdot OP \cdot PQ=\frac{1}{2}\cdot 30\cdot \frac{25}{2}$, since $\angle OQP$ is right. Thus the total area that works is $30\cdot \frac{25}{2}=375$, and the area of the locus of all centers of any circle with radius 1 is $34\cdot 14=442$. Hence, the desired probability is $\frac{375}{442}$, and our answer is $\boxed {817}$.
[/hide]

These problems are so easy yet so annoying  :wallbash: .
Set it up on a coordinate system, draw the outer rectangle with corner at (0,0) and the inner rectangle 1 apart from each side.
The diagonal has equation $ y\equal{}\frac{5}{12}x$. Now draw a line that is a perpendicular distance of 1 above this line, parallel to it.
We need to find the vertical distance between the lines. Draw a perpendicular of length 1 from one line to the other. Now draw a vertical segment from one of the lines to the other that intersects at the same place the perpendicular line does. Since the slope of both lines is 5/12, by similarity,  we can find the vertical distance between the lines is 13/12.
So we find the equation of this line is $ y\equal{}\frac{5}{12}x\plus{}\frac{13}{12}$ and it intersects the inner rectangle at $ (1,3/2)$
Now we want to find the area of the triangle that this line makes with the inner rectangle, which has a corner at $ (1,14)$. By similarity, this triangle has its width as $ 12/5$ its height. So the shorter side is $ 14\minus{}3/2\equal{}25/2$. So the area of the triangle is $ ((25/2)^2\cdot12/5)/2$. The area of half the rectangle is $ 34\cdot13/2$. Dividing, we have $ 375/442$ as the answer.

Sorry for the revive but I found a fast solution:
[hide=Solution]WLOG $AB=15$ and $BC=36$. The total possible region is $13 \cdot 34 \implies 442$. The successful region is figured out by considering WLOG $\triangle{ABC}$ since $\triangle{ABC}$ and $\triangle{ADC}$ are symmetrical in this aspect. The successful area is just a triangle formed by the path of the incenter and circle of radius $1$ rolls around the edge of the triangle. This area is just $\frac{25}{36}$ of the triangle so it's $15 \cdot 36 \cdot \frac{25}{36} \implies 375$ so the answer is $\boxed{\frac{375}{442}} \implies \boxed{817}$. Also the triangle formed by the path of the circle of radius $1$ and $\triangle{ABC}$ have the same incenter and are similar. :)[/hide]

Wow, I cord bashed and actually got the right answer! :D


Let $ a,b,c\geq 0$, no two of which are zero.Prove that:
$ \frac{1}{a^2\minus{}ab\plus{}b^2}\plus{}\frac{1}{b^2\minus{}bc\plus{}c^2}\plus{}\frac{1}{c^2\minus{}ca\plus{}a^2}\geq \frac{3}{ab\plus{}bc\plus{}ca}$

No one likes to do your ineq :-/ .Have you knew that :-/
Assume :$ c \equal{} min(a,b,c)$
$ LHS \ge \frac {1}{a^2 \minus{} ab \plus{} b^2} \plus{} \frac {1}{a^2} \plus{} \frac {1}{b^2}$
$ RHS \le \frac {3}{ab}$
<=>$ (a \minus{} b)^4 \ge 0$


Among nine similar coins, one is counterfeit and lighter than the others. Using a balance only twice, how can the counterfeit coin be identified?

[hide] put six coins on if one side is lighter keep those three, if they are equal, keep the three you left out.

then with the three you have left put 2 on, if ones lighter then its counterfeit, if they are even the one laft over is counterfeit.[/hide]

haha that was kinda fun

Actually, if you have $3^n$ coins, you can figure it out with $n$ balances.  As long as $n$ is a natural number, that is.


algebra
polynomial
linear algebra
matrix

if
$ \sum_{r\equal{}1} ^{n} r\equal{}\frac{1}{2}\times n\times({n\plus{}1})$
$ \sum_{r\equal{}1} ^{n} r^2\equal{}\frac{1}{6}\times n\times({n\plus{}1})(2n\plus{}1)$
$ \sum_{r\equal{}1} ^{n} r^3\equal{}\frac{1}{4}\times n^2\times({n\plus{}1})^2$

what is
$ \sum_{r\equal{}1} ^{n} r^k\equal{}?$

See [url]http://www.mathpages.com/home/kmath147.htm[/url]. I don't know how to prove it without calculus.

[hide="How to compute the formula given k"]Suppose the answer is $ f(x)$. Because the difference $ f(x \plus{} 1) \minus{} f(x) \equal{} x^k$ is a polynomial of degree $ k$, $ f(x)$ is a polynomial of degree $ k \plus{} 1$. Let $ f(x) \equal{} a_{k \plus{} 1}x^{k \plus{} 1} \plus{} a_kx^k \plus{} \ldots \plus{} a_1x$ (notice no constant term because $ f(0) \equal{} 0$). Plugging this into the difference equation above gets you a big polynomial of degree $ k$. Equating coefficients of $ x^0, x^1, \ldots x^k$ gets you a system of $ k \plus{} 1$ linear equations with $ k \plus{} 1$ variables to solve for.

EDIT: For those who know linear algebra and are wondering about there being redundant equations, the coefficient matrix is always upper triangular here.[/hide]

We can just use a recursion...

$ (x\plus{}1)^k\minus{}x^k\equal{}k x^{k\minus{}1}\plus{}\binom{k}{2}x^{k\minus{}2}\plus{}...\plus{}1$

Then sum these for $ x\equal{}0,1,2,..,r\minus{}1$.

The LHS telescopes. The RHS we can use the sum formulas for $ k\equal{}0,1,...,k\minus{}2$, and we are left with the sum for $ k\minus{}1$

Altheman, I'm not exactly sure what you have in mind. (EDIT: I think I see it now, but that doesn't seem terribly useful... but perhaps related to what I wrote below)

Anyway, something I've noticed:
$ \sum_{k \equal{} 1}^n k \equal{} \frac {n(n \plus{} 1)}{2}$
$ \sum_{k \equal{} 1}^n k(k \minus{} 1) \equal{} \frac {(n \plus{} 1)n(n \minus{} 1)}{3}$
$ \sum_{k \equal{} 1}^n k(k \minus{} 1)(k \minus{} 2) \equal{} \frac {(n \plus{} 1)n(n \minus{} 1)(n \minus{} 2)}{4}$

That's all i've bothered verifying so far, but the pattern is rather suggestive: $ \sum_{k \equal{} 1}^n \prod_{i \equal{} 0}^{r \minus{} 1} k \minus{} i \equal{} \frac {1}{r \plus{} 1}\prod_{i \equal{} 0}^r n \minus{} i \plus{} 1$.  Expanding the LHS's of such identities gives equations with the desired sums, so it becomes a matter of starting with $ r \equal{} 1$ (sum of the first n natural numbers) and working up from there.

I don't think there is a general formula, but this definitely suggests a general method.  I believe any formula for $ \sum_{k \equal{} 1}^n k^r$ for a specific $ r$ can be proved by induction.

You are [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=178381]correct[/url] about the pattern, although incorrect about there not being a general formula.  The general formula derived from that method depends on the Stirling numbers of the second kind, whereas the usual [url=http://mathworld.wolfram.com/FaulhabersFormula.html]Faulhaber's formula[/url] depends on the Bernoulli numbers.


[color=darkred]Let $\triangle ABC$with the incircle $C(I)$ and the $A$- exincircle $C(I_{a})$ a.s.o.

Denote the intersections $D\in AI\cap BC$, $E\in BI\cap CA$, $F\in CI\cap AB$.

Prove that $\boxed{\ \left(\forall\right) X\ ,\ \overrightarrow{XI_{a}}\cdot \overrightarrow{EF}+\overrightarrow{XI_{b}}\cdot\overrightarrow{FD}+\overrightarrow{XI_{c}}\cdot\overrightarrow{DE}=0\ }$.

[b][u]Consequence.[/u][/b] $\sum \overrightarrow{GI_{a}}\cdot \overrightarrow{EF}=0$, where $G$- the centroid of $\triangle ABC$.[/color]

Let $O$ is the circumcenter of triangle $ABC$,use [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=139240]this problem[/url] :D , we have $\vec{OI_{a}}.\vec{EF}=\vec{OI_{b}}.\vec{FD}= \vec{OI_{c}}.\vec{DE}= 0$ and of course $\vec{XO}(\vec{EF}+\vec{FD}+\vec{DE}) = 0$ so $\vec{XI_{a}}.\vec{EF}+\vec{XI_{b}}.\vec{FD}+\vec{XI_{c}}.\vec{DE}= 0$


Functional Analysis
real analysis
real analysis unsolved

Prove that the normed space $ (X, \parallel{}.\parallel{})$ is Banach space iff every absolutely convergent sequence in $ X$ implies the convergent of sequence by itself in $ X$

First suppose that $ X$ is a Banach space and $ \sum \parallel{}x_n\parallel{}$ is convergent ,
Let $ S_n = \sum_{i = 1}^{n} x_i$ ,then for $ m > n$ we have $ \parallel{}S_m - S_n\parallel{}\leq\sum_{i = n + 1}^{\infty} \parallel{}x_i\parallel{}$ so $ S_m$ is a Couchy sequence ,since $ X$ is Banach therefore $ S_m$ is cnvergent .

[color=red]Converse:[/color]suppose $ x_n$ is a Couchy sequnce ,we chose $ n_1 < n_2 < ...$ such that

$ \parallel{}x_m - x_n\parallel{}\leq1/2^j$ for all $ m,n\geq n_j$ ,put $ y_{k + 1} = \parallel{}x{}_{n{}_k} - x{}_{n{}_k - 1}\parallel{}$ for $ k\geq 1$ and $ y_1 = x{}_{n{}_1}$

so $ y_k$ is a convergent sequence and by hypothesis we have $ x{}_n{}_{k} = \sum_{i = 1}^{k + 1} y_i$ is cnvergent therefore  $ x_n$ is cnvergent .

[post resized, quote removed by moderator]


function
calculus
derivative
calculus computations

Consider a function $ g(x)$ which is differentiable on some interval $ (\minus{}R,R)$, where $ R>0$ . Given that it satisfies the identity $ xg'(x)\minus{}Ng(x)\equal{}x^N$ for some positive integer $ N$. Prove that $ g(x)$ cannot be analytic at 0 by considering the identity.

Can someone please help. Thank you very much. :P

If $ g(x) \equal{} \sum c_n x^n$, then $ xg' \minus{} Ng \equal{} \sum c_n(n \minus{} N)x^n$. This can't be equal to $ x^N$.

Actually, $ g$ cannot be even $ N$ times differentiable at the origin. If it were, the $ N$th derivative of $ xg'\minus{}Ng$ at $ x\equal{}0$ would be equal to $ 0$, not to $ N!$.


Let $ p>3$ be a prime number. Show that the congruence $ x^2\plus{}3\equal{}0 modp$ has asolution if and only if $ p\equal{}6k\plus{}1$

Use quadric reside :
$ (\frac{\minus{}3}{p})\equal{}1$


geometry
parallelogram
circumcircle
perpendicular bisector
geometry proposed

Let A, C be the unchanged points in the plane. K is the midpoint of AC. B and D are the moving ponts in plane such that B, K, D are collinear and KB=KD. The bisector of angle BCD intersects AB, AD at I, J.
(ABD) intersects (AIJ) at M (M is different from A). Find the locus of M. :blush:

u  want the  answer  as   an equation  or description?

∠AIJ=∠AJI=∠ICD=1/2∠BCD   =>  AI=AJ  ∠O2ID=∠O2AB=180-∠O2AI=180-∠O2JA
Notice ID=DC=AB   O2A=O2J   =>  △O2AB≌△O2JD   =>  ∠ADO2=∠ABO2
=> O2 lies on ⊙O1    BO2=BD   =>  ∠O2DB=∠O2BD=180-∠IAO2=180-∠O2MD
 => △O2JD≌△O2MD  =>  DM=DJ=DC
∠O2BI=∠O2BM  => △O2IB≌△O2MB  => BM=BI=BC
∠CMA=∠CMB+∠AMB=∠DBM+∠BMC=90
So M lies on the semicircle with diameter AC

$ ABCD$ is parallelogram, $ (O_1), (O_2)$ circumcircles of $ \triangle BCD, \triangle AIJ$ and $ (K)$ circle with diameter $ AC.$ $ O_1$ is on perpendicular bisector of $ BD$ through $ K.$ $ \triangle IBC$ is isosceles with $ BC \equal{} BI,$ because $ \angle BCI \equal{} 90^\circ \minus{} \frac {_1}{^2}\angle IBC$ $ \Longrightarrow$ $ \triangle AIJ \sim \triangle BIC$ is isosceles with $ AI \equal{} AJ.$ If $ U, V$ are midpoints of $ AI, AJ,$ then $ O_2U \equal{} O_2V$ and $ BU \equal{} DV,$ the right $ \triangle BO_2U \cong \triangle DO_2V$ are congruent by SAS $ \Longrightarrow$ $ O_2B \equal{} O_2D$ $ \Longrightarrow$ $ O_2$ is also on perpendicular bisector of $ BD$ $ \Longrightarrow$ centers of $ (K), (O_1), (O_2)$ intersecting at $ A$ are collinear $ \Longrightarrow$ they are coaxal $ \Longrightarrow$ $ M \in (K).$


linear algebra
matrix
superior algebra
superior algebra theorems

Let A be an nxn matrix,B be an nxr matrix,C be an rxn matrix and D be an rxr matrix,all over a commutative ring  R  with identity.Suppose there is a matrix  E over  R  such  that  DE=C.Prove  that

                                                  det[A    B]=det(A-BE)detD.
                                                           C    D

What  conclusion  can be drawn if  there  is  a matrix  F  over  R  such  that  FD=B?

$\left( \begin{array}{cc}   A & B \\   DE & D \\ \end{array}\right)=\left( \begin{array}{cc}   I & B \\   0 & D \\ \end{array}\right)\cdot \left( \begin{array}{cc}   A-BE & 0 \\   E & I \\ \end{array}\right)$
$\left( \begin{array}{cc}   A & FD \\   C & D \\ \end{array}\right)=\left( \begin{array}{cc}   A-FC & F \\   0 & I \\ \end{array}\right)\cdot \left( \begin{array}{cc}   I & 0 \\   C & D \\ \end{array}\right)$

Now apply Laplace.


A small hose fills a swimming pool in 16 hours.  A large hose fills the same pool in 12 hours.  With the pool empty, the owner turns on the smaller hose at 8:00 AM.  He turns on the larger hose at 10:00 AM.  Both hoses are used from 10:00 AM to 3:00 PM.   Rounded to the nearest percent, how full is the pool at 3:00 PM?
A.27%	B.48%	C.73%	D. 85%  E.102%

[hide] The small hose can fill $ \frac{1}{16}$ of the pool in one hour. Hence, in two hours it filled $ \frac18$ of the pool. Now the two combined take $ \dfrac{1}{\frac{1}{12}\plus{}\frac{1}{16}}\equal{}\frac{48}{7}$ hours to fill the pool. Thus, in five hours they will fill $ \frac{35}{48}$ of the pool. Thus, $ \frac18\plus{}\frac{35}{48}\approx.85$, so the answer is D.[/hide]

[hide="answer"]From 8:00 to 3:00 is 7 hours. From 10:00 to 3:00 is 5 hours. Therefore the small hose is in the pool for 7 hours and the large hose is in the pool for 5. The small hose fills the pool in 16 hours, so in one hour it fills 1/16 of the pool. The large hose fills the pool in 12 hours, so in one hour it fills 1/12 of the pool. Therefore the pool is filled (7/16)+(5/12), or (21+20)/48, of the way. This is equivalent to (4100/48)%, which, rounded to the nearest one, is 85%.[/hide]


geometry
rectangle
combinatorics unsolved
combinatorics

Is it possible to tile a $6$x$6$ board with $2$x$1$ rectangles such that each vertical and horizontal (without boundary lines of course) is cut at least once?

Any generalization?   ;)

[hide]Assume a given line dividing the board into a A-by-6 portion and a (6-A)-by-6 portion is overlapped by N tiles. The portions will have 6A-N and 36-6A-N remaining squares, respectively. Both these numbers have to be even, hence N is even. Thus, having to cover a line at least once means having to cover it at least twice. There are 5 lines in each direction, totalling 10; twice that would be 20 but there are only 18 tiles. Therefore it is impossible.

An 8x8 board could ( :?: ) be possible under similar rules, as only 28 of the 32 tiles would have to cover a line.[/hide]


function
induction
algebra
functional equation
algebra proposed

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

[quote="Pirkuliyev Rovsen"]Determine all functions $ f\colon \mathbb{Q}\to\mathbb{R}$ such that
$ f(x \plus{} y) \plus{} f(x \minus{} y) \equal{} f(x) \plus{} f(y) \plus{} f(f(x) \minus{} f(y))$
$ \forall x,y \in Q$

_____________________________
Azerbaijan Land of Fire  :ninja:[/quote]

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

$ P(0,0) \iff 2f(0)\equal{}3f(0) \implies f(0)\equal{}0$

$ P(x,0) \iff f(x)\equal{}f(f(x))\ \forall x \in \mathbb{Q}$

So $ f(\mathbb{Q}) \subseteq \mathbb{Q}$ and $ \forall x \in f(\mathbb{Q})\ \implies f(x)\equal{}x$

Let $ x,y \in f(\mathbb{Q})$ then $ P(x,y)$ $ \implies f(x\plus{}y)\equal{}f(x)\plus{}f(y)$ that is cauchy functional equation

The solutions are $ f(x)\equal{}0$ and $ f(x)\equal{}cx\ c,x \in \mathbb{Q}$ and for both we can consider that $ f(\mathbb{Q})\equal{}\mathbb{Q}$

Furthermore if $ f(x)\equal{}cx$ then $ f(f(x))\equal{}f(x) \implies c^2x\equal{}cx\ \implies c\equal{}1$ since $ c\equal{}0$ is already included in the $ f(x)\equal{}0$ solution

So $ \forall x \in \mathbb{Q}$ we have $ f(x)\equal{}0$ and $ f(x)\equal{}x$ that are indeed solution of the equation

[quote="prester"] Let $ P(x,y)$ the assertion $ f(x \plus{} y) \plus{} f(x \minus{} y) \equal{} f(x) \plus{} f(y) \plus{} f(f(x) \minus{} f(y))$

$ P(0,0) \iff 2f(0) \equal{} 3f(0) \implies f(0) \equal{} 0$

$ P(x,0) \iff f(x) \equal{} f(f(x))\ \forall x \in \mathbb{Q}$

So $ f(\mathbb{Q}) \subseteq \mathbb{Q}$ and $ \forall x \in f(\mathbb{Q})\ \implies f(x) \equal{} x$

Let $ x,y \in f(\mathbb{Q})$ then $ P(x,y)$ $ \implies f(x \plus{} y) \equal{} f(x) \plus{} f(y)$ that is cauchy functional equation

The solutions are $ f(x) \equal{} 0$ and $ f(x) \equal{} cx\ c,x \in \mathbb{Q}$ [/quote]

You cant use this conclusion since $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)$ is currently only true for $ x,y \in f(\mathbb{Q})$, not for $ x,y \in \mathbb{Q}$. So you cant use the normal conclusion.

For example, the function $ f(x)\equal{}2\left[\frac x2\right]$ is such that $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)$ $ \forall x,y \in f(\mathbb{Q})$  and is neither $ 0$, neither $ cx$

[quote="pco"][quote="prester"] Let $ P(x,y)$ the assertion $ f(x \plus{} y) \plus{} f(x \minus{} y) \equal{} f(x) \plus{} f(y) \plus{} f(f(x) \minus{} f(y))$

$ P(0,0) \iff 2f(0) \equal{} 3f(0) \implies f(0) \equal{} 0$

$ P(x,0) \iff f(x) \equal{} f(f(x))\ \forall x \in \mathbb{Q}$

So $ f(\mathbb{Q}) \subseteq \mathbb{Q}$ and $ \forall x \in f(\mathbb{Q})\ \implies f(x) \equal{} x$

Let $ x,y \in f(\mathbb{Q})$ then $ P(x,y)$ $ \implies f(x \plus{} y) \equal{} f(x) \plus{} f(y)$ that is cauchy functional equation

The solutions are $ f(x) \equal{} 0$ and $ f(x) \equal{} cx\ c,x \in \mathbb{Q}$ [/quote]

You cant use this conclusion since $ f(x \plus{} y) \equal{} f(x) \plus{} f(y)$ is currently only true for $ x,y \in f(\mathbb{Q})$, not for $ x,y \in \mathbb{Q}$. So you cant use the normal conclusion.

For example, the function $ f(x) \equal{} 2\left[\frac x2\right]$ is such that $ f(x \plus{} y) \equal{} f(x) \plus{} f(y)$ $ \forall x,y \in f(\mathbb{Q})$  and is neither $ 0$, neither $ cx$[/quote]

I supposed to be wrong... :blush:

[quote="Pirkuliyev Rovsen"]Determine all functions $ f\colon \mathbb{Q}\to\mathbb{R}$ such that
$ f(x \plus{} y) \plus{} f(x \minus{} y) \equal{} f(x) \plus{} f(y) \plus{} f(f(x) \minus{} f(y))$
$ \forall x,y \in Q$[/quote]

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

$ P(0,0)$ $ \implies$ $ f(0) \equal{} 0$
$ P(x,0)$ $ \implies$ $ f(f(x)) \equal{} f(x)$

Then  :
$ P(x,x)$ $ \implies$ $ f(2x) \equal{} 2f(x)$
$ P(2x,x)$ $ \implies$ $ f(3x) \equal{} 3f(x)$
And an easy induction (using $ P(nx,x)$) shows that $ f(nx) \equal{} nf(x)$ $ \forall n\in \mathbb N$ and so $ f(ux) \equal{} uf(x)$ $ \forall u\in\mathbb Q^ \plus{}$

So $ f(x) \equal{} ax$ $ \forall x\in\mathbb Q^ \plus{}$ and $ f(x) \equal{} \minus{} bx$ $ \forall x\in\mathbb Q^ \minus{}$ (where $ a \equal{} f(1)$ and $ b \equal{} f( \minus{} 1)$. Then :

$ P(1, \minus{} 1)$ $ \implies$ $ 2a \equal{} a \plus{} b \plus{} f(a \minus{} b)$ and so $ f(a \minus{} b) \equal{} a \minus{} b$
$ P( \minus{} 1,1)$ $ \implies$ $ 2b \equal{} b \plus{} a \plus{} f(b \minus{} a)$ and so $ f(b \minus{} a) \equal{} b \minus{} a$
If $ a > b$ : the first line implies $ a(a \minus{} b) \equal{} a \minus{} b$ and so $ a \equal{} 1$ and the second implies $ \minus{} b(b \minus{} a) \equal{} b \minus{} a$ and so $ b \equal{} \minus{} 1$, and we get the solution $ f(x) \equal{} x$, which indeed is a solution.
If $ a < b$ : the first line implies $ \minus{} b(a \minus{} b) \equal{} a \minus{} b$ and so $ b \equal{} \minus{} 1$ and the second implies $ a(b \minus{} a) \equal{} b \minus{} a$ and so $ a \equal{} 1$, impossible since $ a < b$
If $ a \equal{} b$, we get $ f(x) \equal{} a|x|$ and $ P(1,0)$ $ \implies$ $ a \equal{} a|a|$ and so $ a\in\{ \minus{} 1,0,1\}$ which, indeed, are solutions.

Hence $ 4$ solutions to this equation:
$ f(x) \equal{} 0$
$ f(x) \equal{} x$
$ f(x) \equal{} |x|$
$ f(x) \equal{} \minus{} |x|$


Eric plans to compete in a triathlon.  He can average $ 2$ miles per hour in the $ \tfrac{1}{4}$-mile swim and $ 6$ miles per hour in the $ 3$-mile run.  His goal is to finish the triathlon in $ 2$ hours.  To accomplish his goal what must his average speed, in miles per hour, be for the $ 15$-mile bicycle ride?

$ \textbf{(A)}\ \frac{120}{11} \qquad
\textbf{(B)}\ 11 \qquad
\textbf{(C)}\ \frac{56}{5} \qquad
\textbf{(D)}\ \frac{45}{4} \qquad
\textbf{(E)}\ 12$

the first part will take them 1/8 of an hour, the next will take 1/2. they have 11/8 hours left to travel 15 miles, for an answer of $ \frac{120}{11}$, choice A.


I can't save any HTML and PDF page.Will U correct it?

As I said, use either [url=http://www.getfirefox.com]Firefox 1.5[/url] or [url=http://www.opera.com]Opera 9[/url] browsers and you will have no problems.

Or, alternatively, use Save As ... and save the file in Web Archive .mht format in IE 6.

The PDF links are indeed a bug, to be corrected.

I updated my IE to IE 7 now i don't have this problem.

why don't U have iran flag in your signature?


calculus
integration
trigonometry
analytic geometry
calculus computations

Find $ \iint_{x^2 \plus{} y^2\leq R^2} \frac {R}{\sqrt {R^2 \minus{} x^2}}\,dx\,dy$.

You're integrating inside a disk of radius $ R$, so your integral is
$ \int_{\minus{}R}^{R}\int_{\minus{}\sqrt{R^{2}\minus{}x^{2}}}^{\sqrt{R^{2}\minus{}x^{2}}}\frac{R}{\sqrt{R^{2}\minus{}x^{2}}}\,dx\,dy$
Now pull out $ R^{2}$ in the denominator, so your integrand is
$ \frac{1}{\sqrt{1\minus{}\left(\frac{x}{R}\right)^{2}}}$
Now use the standard substitution
$ \sin\theta\equal{}\frac{x}{R}\Rightarrow R\cos\theta\,d\theta\equal{}dx$

Can we calculate the integral by using the polar coordinate transformation?

[quote="JRav"]You're integrating inside a disk of radius $ R$, so your integral is
$ \int_{ \minus{} R}^{R}\int_{ \minus{} \sqrt {R^{2} \minus{} x^{2}}}^{\sqrt {R^{2} \minus{} x^{2}}}\frac {R}{\sqrt {R^{2} \minus{} x^{2}}}\,dx\,dy$[/quote]

I think it must be $ \int_{\minus{}R}^{R}\int_{\minus{}\sqrt{R^{2}\minus{}x^{2}}}^{\sqrt{R^{2}\minus{}x^{2}}}\frac{R}{\sqrt{R^{2}\minus{}x^{2}}}\,dy\,dx$, hence we have $ \int_{\minus{}R}^{R}\int_{\minus{}\sqrt{R^{2}\minus{}x^{2}}}^{\sqrt{R^{2}\minus{}x^{2}}}\frac{R}{\sqrt{R^{2}\minus{}x^{2}}}\,dy\,dx \equal{} \int_{\minus{}R}^{R}2R\,dx\equal{} 4R^2$

[quote="sos440"][quote="JRav"]You're integrating inside a disk of radius $ R$, so your integral is
$ \int_{ \minus{} R}^{R}\int_{ \minus{} \sqrt {R^{2} \minus{} x^{2}}}^{\sqrt {R^{2} \minus{} x^{2}}}\frac {R}{\sqrt {R^{2} \minus{} x^{2}}}\,dx\,dy$[/quote]

I think it must be $ \int_{ \minus{} R}^{R}\int_{ \minus{} \sqrt {R^{2} \minus{} x^{2}}}^{\sqrt {R^{2} \minus{} x^{2}}}\frac {R}{\sqrt {R^{2} \minus{} x^{2}}}\,dy\,dx$, hence we have $ \int_{ \minus{} R}^{R}\int_{ \minus{} \sqrt {R^{2} \minus{} x^{2}}}^{\sqrt {R^{2} \minus{} x^{2}}}\frac {R}{\sqrt {R^{2} \minus{} x^{2}}}\,dy\,dx \equal{} \int_{ \minus{} R}^{R}2R\,dx \equal{} 4R^2$[/quote]
Year, you are right.


function
algebra
domain
functional equation
algebra unsolved

I'm not sure if this is the right forum for this question ... but here I go:

Find all functions [i]f[/i] : [b]R[/b] --> [b]R[/b] satisfying [i]f[/i] ([i]x[/i]+[i]y[/i]) = [i]f[/i] ([i]xy[/i]) for all real [i]x[/i] and [i]y[/i].

Also: Consider the same problem when the domain is the positive reals and the functional equation holds just for [i]x[/i] > 0, [i]y[/i] > 0.

constant function. just substitute $ y\equal{}0$

[hide]Let $ y\equal{}0$, then $ f(x)\equal{}f(0)$, so $ f(x)\equal{}c$ for some constant $ c$.

If we restrict the problem to $ x,y>0$, then let $ f(4)\equal{}c$.  Suppose $ xy\equal{}4$, then $ x\plus{}y\equal{}x\plus{}4/x$, which takes on all values greater than or equal to 4.  Therefore $ f(x)\equal{}c$ for all $ x\ge 4$.  Now suppose $ x\plus{}y\equal{}4$, then $ xy\equal{}x(4\minus{}x)$, which takes on all values less than or equal to 4.  Therefore $ f(x)\equal{}c$ for some constant $ c$ for all $ x$.[/hide]


trigonometry
calculus
calculus computations

can you show :D

$ \zeta \left( {1 \minus{} z} \right) \equal{} \left( {\frac{z}{{2\pi e}}} \right)^z \sqrt {\frac{{8\pi }}{z}} \;\cos \left( {\frac{{\pi z}}{2}} \right)\;\left(  1 \plus{} O \left( {\frac{1}{z}} \right)  \right)$

( for $ z\in\mathbb C$ )

The O notation implies a limit- do you mean that this applies for $ z\to \infty$ near the positive real axis? It's certainly not global.

form the $ \zeta(z)$ functional equations we have :
$ \zeta(z) \equal{} 2^{z}\pi^{z \minus{} 1}\sin(\pi z /2)\Gamma(1 \minus{} z)\zeta(1 \minus{} z)$
substitute $ z \equal{} 1 \minus{} s$ then we have
$ \zeta(1 \minus{} s) \equal{} \frac {2}{2^{s}}\pi^{s}\cos(\frac {\pi s}{2})\Gamma(s)\zeta(s)$
as my friend Stirling said  $ \Gamma(s \plus{} 1) \approx \frac {s^{s}}{e^{s}}\sqrt {2\pi s}$ Hence
$ \Gamma(s)\approx \frac {s^{s}}{e^{s}} \sqrt {\frac {2\pi}{s}}$ Since $ \Gamma(s \plus{} 1) \equal{} s\Gamma(s)$
and $ \zeta(s) \approx 1 \plus{} O(\frac {1}{s})$ thus we have :
$ \frac {2}{2^{s}}\pi^{s}\cos(\frac {\pi s}{2})\Gamma(s)\zeta(s)\approx \frac {2}{2^{s}}\pi^{s}\cos(\frac {\pi s}{2})\frac {s^{s}}{e^{s}} \sqrt {\frac {2\pi}{s}}\left( 1 \plus{} O(\frac {1}{s}) \right)$
$ \equal{} \left( {\frac {s}{{2\pi e}}} \right)^s \sqrt {\frac {{8\pi }}{s}} \;\cos \left( {\frac {{\pi s}}{2}} \right)\;\left( 1 \plus{} O \left( {\frac {1}{s}} \right) \right)$
All this is true if and only if when
[quote="jmerry"] $ z\to \infty$ near the positive real axis[/quote]


geometry
circumcircle
geometric transformation
reflection
cyclic quadrilateral
geometry unsolved

Dear Mathlinkers,
Let ABC be a triangle, 0 the circumcircle of ABC, 1 the incircle of ABC, I the centre of 1, A' a point on the arc BC which doesn't contain A,
B', C' two points on 0 so that the triangle A'B'C' has 1 for incircle,
X the meet point of the parallel to AA' passing through I with BC, A" the midpoint of the arc BC which doesn't contain A, J' the meet points of A'B' and BC, K' the meet point of A'C' and BC, 2 the circumcircle of A'J'K'.
Prove that A"X pass through the midpoint of the arc J'K' of 2 which doesn't contain A'.
Sincerely
Jean-Louis

Proof is almost identical to another problem posted before: [url]http://www.mathlinks.ro/viewtopic.php?t=209898[/url].

Tangent $ s$ to the incircle $ (I)$ parallel to $ BC$ cuts $ AA'$ at $ P$ and tangent $ t$ to $ (I)$ from $ P$ cuts $ BC$ at $ Q.$ Let $ A^* \in AA'$ be arbitrary outside of $ (I)$ and let tangents to $ (I)$ from $ A^*$ cut $ BC$ at $ J^*, K^*.$ $ (I)$ is either excircle (or incircle) of the $ \triangle A^*J^*K^*.$ $ AA'$ cuts $ BC$ at $ D$ and parallel $ p \parallel PQ$ through $ D$ is tangent to the incircle (or excircle) $ (I^*)$ of the $ \triangle A^*J^*K^*$ centrally similar to $ (I)$ with similarity center $ A^*.$  This incircle (or excircle) is centered on the bisector $ DI^* \parallel PI$ of the angle $ \widehat{p, BC}.$ Let $ M^*$ be the midpoint of the segment $ II^*.$ This is the midpoint of one arc $ J^*K^*$ of the circumcircle $ \odot(A^*J^*K^*)$ and at the same time, circumcenter of the cyclic quadrilateral $ IJ^*I^*K^*$ with circumcircle $ (M^*).$ $ M^*$ is on midparallel $ m$ of $ DI^* \parallel PI.$ Let $ L \in (M^*)$ be reflection of $ I$ in $ m$ and let $ IL$ cut $ BC$ at $ Q'$ Then $ \overline{Q'J^*} \cdot \overline {Q'K^*} \equal{} \overline{Q'I} \cdot \overline {Q'L}.$ Let $ F \equiv s \cap BC$ be the common point at infinity of the 2 parallels $ s \parallel BC.$ When $ A^*$ coincides with $ P \in AA',$ we get $ \overline{Q'F} \cdot \overline {Q'Q} \equal{} \overline{Q'I} \cdot \overline {Q'L} \equal{} \text{const}.$ Since $ \overline{Q'F} \equal{} \pm \infty,$ it follows that $ \overline {Q'Q} \equal{} 0$ and the points $ Q' \equiv Q$ are identical. Let the line $ A'Q$ cut the circumcircle $ \odot(A'IL)$ at $ T.$ Then $ \overline{QA'} \cdot \overline {QT} \equal{} \overline{QI} \cdot \overline {QL} \equal{} \overline{QJ^*} \cdot \overline {QK^*}.$ It follows that $ A'TK^*J^*$ is cyclic for any $ A^* \in AA'.$

First, $ A^* \in AA'$ can be identical with $ A,$ in which case $ J^*, K^*$ coincide with $ B, C,$ so that circumcircle $ (O)$ of the $ \triangle A'BC$ is in the pencil $ \mathcal P \equal{} \left\{\odot (A'TK^*J^*) | A^* \in AA' \right\}.$ Next, when $ A^* \in AA'$ is at infinity, tangents to the incircle $ (I)$ from $ A^*,$ cutting $ BC$ at $ J_0, K_0,$ are parallel to $ AA'.$ The circumcircle $ \odot(A'J_0K_0)$ also belongs to $ \mathcal P.$ The circles $ (O) \equal{} \odot(A'BC), \odot (A'J_0K_0)$ intersect at $ A'$ and, by the problem [url]http://www.mathlinks.ro/viewtopic.php?t=233986[/url], at the tangency point $ T \in (O)$ of the A-mixtilinear incircle of the $ \triangle ABC.$ Finally, $ A^*\in AA'$ can be identical with $ A',$ so that the circumcircle $ \odot(A'J'K')$ also belongs to the pencil $ \mathcal P.$ The midparallel $ m$ of $ DI^* \parallel PI,$ the locus of centers of the circles $ (M^*)$ with diameters $ II^*,$ is clearly perpendicular to $ QI$ and $ A'' \in m,$ as center of of the circle $ (A'')$ with diameter $ II_a$ (where $ (I_a)$ is the A-excircle of the $ \triangle ABC$). If parallel to $ AA'$ through $ I$ cuts $ BC$ at $ X,$ then by the same problem quoted above, $ X \in m.$

Dear Yetti and Mathlinkers,
thank for your interesting proof.
Finally, I found a complete synthetic proof that you will see on my website next.
Sincerely
Jean-Louis


[quote="TRAN THAI HUNG"]Let m,n $ \in$ N ,x $ \in$ [0,1].Show that: :D
$ \(1 - x^n )^m + (1 - (1 - x)^m )^n \ge 1$[/quote]

[hide]Suppose $ x$ is the probability event $ X$ occurs. Then $ (1-x^n)^m$ is the probability that in each of $ m$ occurrences of $ n$ events, $ \exists\neg X$. Also, $ (1-(1-x)^m)^n$ is the probability that in each of $ n$ occurrences of $ m$ events, $ \exists X$. So $ X$ can be the event a unit square on an $ m\times n$ grid is marked. If the statement corresponding to the probability $ (1-x^n)^m$ is not true, then a row of length $ n$ is all marked. But then every column (length $ m$) is marked at least once, so the second statement is true. Similarly, if the statement corresponding to $ (1-(1-x)^m)^n$ is not true, then a column of length $ m$ is all unmarked. But then every row (length $ n$) is not marked at least once. Since at least one of the original statements corresponding to the terms being summed is true, the sum of the probabilities is greater than $ 1$. [/hide]

Nice solution,but I'm still not very clear about it.Can you tell me some more about "using probability to solve problems"or some more examples?Please,thanks a lots. :lol:


Let $ a,b,c>0$ .find the best $ t$ such that

\[ \frac{a^3\plus{}b^3\plus{}c^3}{abc}\plus{}t(\frac{ab\plus{}bc\plus{}ca}{a^2\plus{}b^2\plus{}c^2})\ge t\plus{}3\]

Inequality is equivalent with
\[ \frac {a^3 \plus{} b^3 \plus{} c^3 \minus{} 3abc}{abc}\ge t\frac {a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} bc \minus{} ca}{a^2 \plus{} b^2 \plus{} c^2}
\]
or
\[ \frac {a \plus{} b \plus{} c}{abc}\ge t\frac {1}{a^2 \plus{} b^2 \plus{} c^2}
\]
But
\[ \frac {(a^2 \plus{} b^2 \plus{} c^2)(ab \plus{} bc \plus{} ca)}{abc}\ge 9
,\]
then t=9


http://www.mathlinks.ro/Forum/viewtopic.php?t=25780

  darij


inequalities
complex numbers
inequalities proposed

Let $ z_1$ and $ z_2$ be two non-zero complex numbers .Prove that :
\[ \left(\frac{1}{|z_1|} \plus{}\frac{1}{|z_2|}\right)\left(|\sqrt{z_1 z_2} \plus{}\frac{z_1\plus{}z_2}{2} |\plus{}|\sqrt{z_1 z_2} \minus{}\frac{z_1\plus{}z_2}{2}| \right) \geq 4\]


[quote="silver1989"]Let a,b,c$\geq 0$ such  that a+b+c=3.Prove that:
 $\sum\frac{1}{6-ab}\leq \frac{3}{5}$[/quote]
Let $a=3x,$ $b=3y,$ $c=3z$ and look here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=114583


[color=darkred]Determine the sequence $ (a_n)$, $ a_i\in \mathbb{N^*}$ for which exist natural numbers $ p$ and $ q$ such that

1) $ \sum^n_{i \equal{} 1} a_i^2 \equal{} p^3$;

2) $ \sum^n_{i \equal{} 1} a_i^3 \equal{} q^2$.[/color]

For example $ n\equal{}m^6,a_i\equal{}k^6$.


I assume this is standard snare drumming?


Here's a mathematical jewel; I'm not sure where it originates from, but it's great exercise:

[b]Problem. [/b] Let $(A_i,B_i)_{1 \leq i \leq h}$ be a finite family of subsets of the integers such that, for all $i$, $|A_i| \leq k, |B_i| \leq l, A_i \cap B_i = \emptyset$, and for all $i \neq j$, either $A_i \cap B_j \neq \emptyset$, or $A_j \cap B_i \neq \emptyset$ (or both). Prove that $h \leq \frac{(k+l)^{k+l}}{k^kl^l}$.

Enjoy!

--Vesselin

What if we decide never to use the same integer twice, and make the family as big or small as we like?

e.g. $k=l=1$, $A_i=\{2i+1\}, B_i=\{2i\}$.
Surely I have missed something here, but I can't fathom what it is...

Sorry, I intended to write

'$A_i \cap B_j \neq \emptyset$ or $A_j \cap B_i \neq \emptyset$'

instead of

'$A_i \cap B_j  = \emptyset$ or $A_j \cap B_i = \emptyset$.'

In other words, we have $A_i \cap B_i = \emptyset$, but $(A_i \cap B_j) \cup (A_j \cap B_i) \neq \emptyset$ for $i \neq j$. I've edited the above post.

--Vesselin


for $ A\equal{}\left( \begin{array}{cccc}30&20&15&12\\20&15&12&15\\15&12&15&20\\12&15&20&30\end{array}\right)$
and $ B\equal{}\left(\begin{array}{cccc}x&a&a&a\\\minus{}a&x&a&a\\\minus{}a&\minus{}a&x&a\\\minus{}a&\minus{}a&\minus{}a&x\end{array}\right)$
caculate detA and detB
I caculated them and found the answer because they're only $ 4\times 4$ matrices but I wish a nice solution. can you help me, please? :D

I'm not going to go through all of the calculations but often times, especially for large matrices, it's a lot easier to turn your matrix into an upper or lower triangular by means of row and column operations.  Recall that a determinant is not changed if you add a multiple of one row or column to another.


I've been working on this for a bit and got *close*, but not quite there
Evaluate sec20 + sec40 + sec80 (degrees)
If you write it all in terms of cos, you can use double angle formulas to find the denomitaor (which leads to cos20cos40cos80=1/8), if you'll use that in your solution.  Help please?


?? :?  :?

fekr nakonam kheyli sakht basheh...

[quote="jensen"][quote="amirhtlusa"]donbaleh ye $a_n$ be soorat e zir tarif shodeh ast:

$\forall k \exists i$ s.t. $a_i>0$
$\forall k \exists j$ s.t. $a_j<0$
$\forall i\in\mathbb{Z^+}, a_i-a_j|i-j$

saabet konid har addad e mosbat daghighan yek bar dar $\{a_n\}$ zaher mishavad.[/quote]
$\forall k \exists i$ s.t. $a_i>0$
$\forall k \exists j$ s.t. $a_j<0$
pas $k$ inja chi karast?[/quote]

$\forall k \exists i>k$ s.t. $a_i>0$
$\forall k \exists j<k$ s.t. $a_j<0$ :)


algebra
difference of squares
special factorizations

05 State #21.  How many positive three-digit integers with each digit greater                    than $ 4$ are divisible by $ 6$? (looking for an easier way to approach this....)

08 State #29.  Given that $ x^2\minus{}5x\plus{}8\equal{}1$, what is the positive value of $ x^4\minus{}10x^3\plus{}25x^2\minus{}9$?

00 State #30.  Joe bought a pumpkin that cost $ 10$ cents more per pound than his sister's. Together, the two pumpkins weighed $ 20$ pounds, but Joe's pumpkin was heavier. Joe paid $ \ 3.52$ dollars and his sister paid $ 48$ cents. How many pounds did Joe's pumpkin weigh?

04 State #17.  What is the sum of the tens digit and the units digit in the decimal representation of $ \ {9}^{2004}$?

And.. Any suggestions as to what formulas I should memorize, or maybe what website could be useful for upcoming State Competitions??  :)

[hide="hint for 29"]Factoring $ x^4\minus{}10x^3\plus{}25x^2\minus{}9$ yields $ (x^2\minus{}5x\minus{}3)(x^2\minus{}5x\plus{}3)$.[/hide]

[hide="Hint for 2004 State Sprint #17"]Find a pattern with the first few.[/hide]

[hide="21."]
Divisibility of 6 => Divisibility of 3, 2.
Then we break up the parts into the first two digits and the units digit.
Unit digit must be 6 or 8.
If 6, then the sum of other two digits must be divisible by 3.
If 8, then the sum of other two digits must be 1 more than a number divisible by 3.

10: 1
11: 2
12: 3
13: 4
14: 5
15: 4
16: 3
17: 2
18: 1
Ans: 1+3+4+4+3+1=[b]16[/b]

[/hide]

ok,for 29(i kno that others have posted this but...) first look at x^4-10x^3+25x^2-9...so lets group this into two parts:(x^4-10x^3+25x^2)-9,which is x^2(x^2-10x+25)-9,which equals [(x)(x-5)]^2-3^2,so this is the difference of squares so (x^2-5x-3)(x^2-5x+3)...the other equation they gave you was x^2-5x+8=1,which is equal to x^2-5x=-7,so the equation we just solved for is equal to (-7-3)(-7+3),so (-10)(-4) or 40!!

hope it helped! :D

For 04 state #17
look for a pattern for the patterns of 9 powered
81,29.26 and so on till you find the pattern


Hi i need help with this problem. If someone could please explain it to me:

The sum of the digits of a two-digit number is subtracted from the number. The units digit of the result is 6. How many two-digit numbers have this property?

I already came up with:
15
24
42
51
33
60
06

I heard there were three more answers but cant think of them. Please help.

first, 06 is not a 2-digit number
[hide="solution"]
let this 2 digit number be $10a+b$, where $a$ and $b$ are the tens and units digits, respectively
the sum of the digits is $a+b$
so the new number is $9a$
so $9a$ must end in 6
this can only happen if $a=4$
so any case where $a=4$ works
and the solutions are 40, 41, 42, 43, 44, 45, 46, 47, 48, 49
you said there were supposed to be ten answers
here they are[/hide]

wow i had no idea that it was that complicated- thanks alot for ur help


Call a set a [i]square set[/i] iff it contains exactly four positive integers $a,b,x,y$ such that
\[a^{2}+b^{2}=x^{2}+y^{2}\] and $x+y \neq a+b$. The number of distinct square-sets with no element greater than $2007$ is $N$. Find the remainder when $N$ is divided by $1000$.

(This problem was given to me by a friend, who said that there was a nice solution, but won't tell me the solution. I don't see it, though. I just see lots of factorizing integers. Maybe a recursion?)

Pretty simple:
$(a^{2}+b^{2})(x^{2}+y^{2}) \ge (ax+by)^{2}$
Also known as Cauchy's inequality.

[quote="t0rajir0u"]When kunny says equality I am pretty sure he means the [i]equality[/i]

$(a^{2}+b^{2})(c^{2}+d^{2}) = (ac+bd)^{2}+(ad-bc)^{2}$[/quote]

That's right.This can be obtained as follows.

$(a^{2}+b^{2})(c^{2}+d^{2})=(a+bi)(a-bi)(c+di)(c-di)$
$=(a+bi)(c-di)(a-bi)(c+di)$
$=(ac+bd)^{2}+(ad-bc)^{2}.$

Similarly we can get $(a^{2}+b^{2})(c^{2}+d^{2})=(ac-bd)^{2}+(ad+bc)^{2}.$