$ x,y,z>0$      and      $ x\plus{}y\plus{}z\equal{}3$

 Prove that \[ \frac{x^3}{y^3\plus{}8}\plus{}\frac{y^3}{z^3\plus{}8}\plus{}\frac{z^3}{x^3\plus{}8}\geq\frac{1}{9}\plus{}\frac{2}{27}(xy\plus{}xz\plus{}yz)\]

[quote="Serod"]$ x,y,z > 0$      and      $ x \plus{} y \plus{} z \equal{} 3$

 Prove that
\[ \frac {x^3}{y^3 \plus{} 8} \plus{} \frac {y^3}{z^3 \plus{} 8} \plus{} \frac {z^3}{x^3 \plus{} 8}\geq\frac {1}{9} \plus{} \frac {2}{27}(xy \plus{} xz \plus{} yz)
\]
[/quote]

By Cauchy Schwarz and Vasc inequality we have
\[ \sum {\frac{{x^3 }}{{y^3  \plus{} 8}}}  \ge \frac{{\left( {\sum {x^2 } } \right)^2 }}{{\sum {xy^3 }  \plus{} 8\sum x }} \ge \frac{{3\left( {\sum {x^2 } } \right)^2 }}{{\left( {\sum {x^2 } } \right)^2  \plus{} 72}} \equal{} \frac{{3\left( {9 \minus{} 2t} \right)^2 }}{{\left( {9 \minus{} 2t} \right)^2  \plus{} 72}}
\]
where $ t \equal{} xy \plus{} yz \plus{} zx \le 3$
Hence it suffices to prove that
\[ \frac{{3\left( {9 \minus{} 2t} \right)^2 }}{{\left( {9 \minus{} 2t} \right)^2  \plus{} 72}} \ge \frac{1}{9} \plus{} \frac{2}{{27}}t
\]

\[ \Leftrightarrow \left( {3 \minus{} t} \right)\left( {4t^2  \minus{} 180t \plus{} 1017} \right) \ge 0
\]

It's done. :)

It's an old problem with Am-Gm:
$ \frac{a^3}{b^3\plus{}8} \plus{} \frac{b\plus{}2}{27} \plus{} \frac{b^2\minus{}2b\plus{}4}{27} \geq\ \frac{a}{3}$
 :)

[quote="Serod"]$ x,y,z > 0$      and      $ x \plus{} y \plus{} z \equal{} 3$

 Prove that
\[ \frac {x^3}{y^3 \plus{} 8} \plus{} \frac {y^3}{z^3 \plus{} 8} \plus{} \frac {z^3}{x^3 \plus{} 8}\geq\frac {1}{9} \plus{} \frac {2}{27}(xy \plus{} xz \plus{} yz)
\]
[/quote]
this is a problem in Iranian National Olympiad (3rd Round) 2008
You can see here:[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=1241795#1241795[/url]
and here:[url]http://www.mathlinks.ro/viewtopic.php?t=223062[/url]

Appears very simple though. Serd, are u an iranian?


Find all those $a,b,c$ whole numbers for wich :
$a^{2}+b^{2}+c^{2}+3 \lneqq ab+3b+2c$

[quote="Thaakisfox"]Find all those $ a,b,c$ whole numbers for wich :
$ a^{2} \plus{} b^{2} \plus{} c^{2} \plus{} 3 \lneqq ab \plus{} 3b \plus{} 2c$[/quote]
My solution:
$ a^2 \plus{} b^2 \plus{} c^2 \plus{} 3 < ab \plus{} 3b \plus{} 2c$
<=>$ (a \minus{} \frac {1}{2}b)^2 \plus{} \frac {3}{4}(b \minus{} 2)^2 \plus{} (c \minus{} 1)^2 \leq \minus{} 1$
 :P


A weather station has 3 rain gauges. One is in the form of a circular cylinder, open at the top, with a base diameter of 20 cm and a height of 40 cm.
The second rain gauge is in the shape of a single circular cone, with its point downwards. The open top of the cone has a diameter of 20 cm and its height is 40 cm.
The third rain gauge is a double cone made from two circular cones on top of each other with their tips touching (like an hour glass). They are both of the same dimensions as the second rain gauge (i.e. this one is in total double the height of the second one).

The 3 rain gauges stand in the open next to each other. After a rainstorm it is noted that the depth of water in the cylindrical rain gauge is 10 mm. What is the depth of water in the other two rain gauges?

[hide]
Since the second rain gauge has 1/3 of the volume of the first, so 30 cm?
The third would be the same.
This looks to be to simple however.
Finding the volumes:
10mm=1cm
$(10)^2(\pi)=100\pi\text{ cm}^3$

Second thing:
$(1/3)(\pi)(10^2)(h)=100\pi$
$(1/3)(100)(h)=100$
$(1/3)(h)=1$
$h=3$
h=30 mm.
Same for last, because it doesnot go into the top, or else I am not visualizing this correctly.[/hide]

Nope, not correct. Remember, in the second cone the water fills from the narrow end upwards, while in the third the cone is turned around so it fills from the big base upwards.
The second one looks like this [size=150]v[/size]
and the third one looks like this [size=150][u]X[/u][/size]

Added a trivia (physics)  question: Why most of the rain gauges are  shaped like the third one? (Or why the third shape is preferable than the other two)

[hide]
It can be shown that the height of the water in the second is 3 mm.
It can also be shown that the height of the water in the third is
forty minus four times the cube root of one-thousand nine-hundred eighty-five halves cm-which is a little larger than 1 mm

If anybody could tell me how to type the cube root sign I would appreciate it.[/hide]

I'm sorry lund, but both your answers are incorrect.

Maybe this problem is too hard for getting started forum? Could some mod please move it?

IMO - The problem is okay for this level.
[hide="Hint "]

1. The  volume of [b]water [/b] is same in all three. (Volume depends only on the area of opening)
2. For the first case,  assuming the height is $h$  the radius of the "cone" of water is  $\frac h 4$ .[/hide]

Do you mean that the depth of the water is 10 cm and not 10 mm?

This is my solution:

[hide]

The first thing we need to find is the volume of rain that fell.

The depth of the water in the cylindrical guage is $10\ mm$ or $1 \ cm$

Knowing that the volume of a cylinder is $\pi r^2 h$ we can find the volume of water which is $\pi(10 \ cm)^2(1 \ cm)$ or $100 \pi \ cm^3$.

Our next task is to find how deep the water would be in the guage that is in the shape of an inverted cone.

My illustration of the cone will help in clarifying my solution.

The volume of a cone is $\frac{1}{3} \pi r^2 h$

From what we have given the ratio of the altitude to the radius of the cone is $4: 1$

let $h$ = the depth of the water in the cone.  So $r = \frac{h}{4}$.

So knowing the volume we can say $100 \pi = \frac{1}{3} \pi (\frac{h}{4})^2h$

$\frac{1}{48} h^3 = 100$. So $h^3 = 4800$ and $h = 16.87 \ cm.$

Finding the volume in the third guage is a little more involved. (See sketch of double cone).

Again, let $h$ = the depth of the water.

$40 - h$ = the distance from the vertex of the guage to the surface of the water.

This time the volume of the water is expressed as follows:

$100 \pi = \frac{1}{3} \pi (10)^2(40) - \frac{1}{3} \pi [\frac{1}{4}(40 - h)]^2(40 - h)$

$300 = 4000- \frac{1}{16}(40 - h)^3$

$(40 - h)^3 = (3700)(16) = 59200$

$40 - h = 38.97$  so $h = 1.026 \ cm$

[/hide]

You may want to check me on that second answer. It doesn't seem right to me.

ritchjp, you sure got the right idea by setting $r=\frac{h}{4}$ :D , but both your answers are still wrong. The volume should be $100\pi{cm^3}$. Also, at the end of your solution, if $\frac{1}{48}h^3=10$ then $h^3=480$, not $\frac{5}{24}$.

Melissa, I see that I should have multiplied instead of divided. (honest mistake).

As for the volume, $10 \ mm$ is $1 \ cm$.

I will edit.

[quote="lund"]If anybody could tell me how to type the cube root sign I would appreciate it.[/quote]

$\$$\sqrt[3]{x}$\$$$\implies \sqrt[3]{x}.$

(enclose it in dollar signs)

Very good, ritchjp, both your answers are correct now :thumbup:  :lol:


We put alternating current in the following circuit. What should the frequency of the current be so that the magnitude of the current in the main branch ($A$) is independent from $R$?

Sorry, I've solved for a wrong circuit :blush: It should be $\omega = \sqrt{2/LC}$. Is that correct?

Ok, here's the solution. The complex impedance is \[z = \frac{j\omega L+\left(R-\frac{j}{\omega c}\right)}{R+j\omega L-j/\omega C},\] where $j^{2}\equiv-1$. The complex current in the main branch is $i = u / z$. We choose $u = U_{0}$ (its phase is 0), so the modulus of the impedance must be independent of $R$, i.e. its derivative with respect to $R$ must be zero at all times. The modulus is \[|z| = \frac{(L/C)^{2}+R^{2}\omega^{2}L^{2}}{R^{2}+(\omega L-1/\omega C)^{2}},\] and taking its derivative and equating it with zero we get $\omega^{2}LC = 2$.


slove for x
           where [x] is the largest integers which is smaller than x

[quote="feliz"]Clearly, x is an integer (it is a difference of integers). We have x + x/6 = x/2 + 2x/3. Then,

[logx] = {x/6} - {x/2} - {2x/3}.

But this is less than one. So [logx] = 0 and logx < 1. Now we have only ten possible values.[/quote]
Thanks you very much!


algebra
polynomial
inequalities
calculus
inequalities unsolved

Well, probably it is somewhere in the forum... But I don't know what name I should use to look for it...

Let $ x_1, x_2, ..., x_n$ positive real numbers and $ a_k \equal{} \sum_{i_1 < i_2 < ... < i_k} x_{i_1}x_{i_2}...x_{i_k}$. Prove that

$ \sqrt [k]{\frac {a_k}{\binom{n}{k}}} \ge \sqrt [k \plus{} 1]{\frac {a_{k \plus{} 1}}{\binom{n}{k \plus{} 1}}}$,

where $ 1 \le k \le n \minus{} 1$.

Thanks in advance.

Thank you. Knowing the inequality's name, it was easy to find a proof...

http://www.artofproblemsolving.com/Wiki/index.php/Maclaurin's_Inequality

By the way, is there another proof? Is there a proof without calculus?


Study if this serie is convergent or not sum(n=1-> infinit) (2^(-1)^n-n).


Find all natural $ (x,y)$ such that
$ y^{x^{2}}\equal{}x^{y\plus{}2}$

[hide="Messy"]

$ (1, 1)$ and $ (2, 2)$ are obvious solutions.
Let $ x^{2}\equal{} z$.  Then
\[ y^{2z}\equal{}z^{y\plus{}2}\]
Clearly, $ p|z\iff p|y$ for a prime $ p$.  Thus, $ (z, y\plus{}2)|2$.
Let $ p^{k}\parallel y$.  Then $ p^{2zk}\parallel z^{y\plus{}2}\implies y\plus{}2|2zk$
If $ (y\plus{}2, z) \equal{} 1$, then $ y\plus{}2|2k$, so $ p^{\frac{y}{2}\plus{}1}$ or $ p^{y\plus{}2}$ divides $ y$. This is clearly impossible. So $ (y\plus{}2, z)\equal{}2$. Then $ \frac{y}{2}\plus{}1|2k$ . We then want to determine when $ p^{\frac{y\plus{}2}{4}}| y$. Clearly, unless $ y\equal{}2$, $ p\neq 2$, because $ y\equiv 2\mod{4}$.  So checking cases,

$ p\equal{}3\implies y\equiv 6\mod{12}$, we find no solutions.
$ p\equal{}5\implies y\equiv 10\mod{20}$, again no solutions.
For larger $ p$, we will have $ p^{\frac{y\plus{}2}{4}}>y$.

Hence, there are no other solutions.[/hide]

A lemma
$ a^{m}\equal{}b^{n}$
$ gcd(m,n)\equal{}d$
Then exist c
$ a\equal{}c^{\frac{n}{d}}$
$ b\equal{}c^{\frac{m}{d}}$


function
calculus
derivative
6th edition

Can somebody post a solution to the first problem.I believe the only possible answer is(1,1) but the case when both x and y are in(1,2) interval was difficult to investigate.

Are you sure there are no other solutions?  Try evaluating ${x^{-y}+y^{-x}+x+y}$ at x = 0.7, y = 0.8, and x = 0.7, y = 0.9.  This means there is a solution x = 0.7, 0.8<y<0.9, I think.  Similarly there are solutions for many other values of x.  This was the question I didn't solve, and so I too would like to see a solution.

If I'm not mistaken, there are an infinity of solutions... but I couldn't find them. anybody found 'em?

I don't quite think so. I know that, because I created this problem. I checked it before. ;)

okay... so let's see how you solved the problem. it ought to be good

Well, it's kind of late and the proof is quite long. Anyway, just give a pm with your e-mail adress and i'll send you the file there. ;)

We all want to see your solution, Cezar. So, post the solution OR the file here.

[quote="perfect_radio"]Did you bother reading the post before mine?

There you could have seen an argument which proves that there is more than a solution :) ( I also found out that discouraging result, and thus I gave up on the problem) We can find many solutions if we take a good $x$, say between 0.7 and 1[/quote]

So what is your $y$? With sure   there is no solution with your $x$ and $y>1$, what is trivial to prove.

If you have a look at the post previous to perfect_radio's (mine), then that shows that there is a solution y with 0.8<y<0.9 when x = 0.7.  If we could have found y explicitly then the problem would have been solved!
Jack

If $f(x,y)=x^{-y}+y^{-x}+x+y-4$ then it is easy to check that $f(2,2)>0$, and $f(1/2,1/2)=2\sqrt{2}-3<0$. It follows that every curve from $(2,2)$ to $(1/2,1/2)$ contains at least one point where $f(x,y)=0$, and therefore the set of solutions is infinite. In the attached pdf file the shaded region is where $f(x,y)>0$. I think one can prove that the set of solutions is the union of two curves $(x,g(x))$ and $(g(y),y)$ where $g$ is a suitable function from $(0,1]$ into itself, but I don't think that $g$ as a simple expression in terms of elementary functions.

Yes - indeed, I found that either $x=y=0$, $x=y=1$ or $0<x<1$ and $0<y<1$. Actually, for all $0<x<1$ there are exactly two solutions for $y$, one satisfying $0<y<x$, the other one $x<y<1$. These things are quite easy to prove. I too doubt that it is possible to express these two solutions in terms of elementary functions in $x$.

Can somebody post full and 100% correrct solution?

I am still waiting on Cezar's solution...

I'm wondering... how will Valentin grade this problem  :?

Peter Scholze, how did you prove that there are exactly 2 solutions? i remember i tried with derivatives, but i got smth weaker, only a solution

I think that Valentin or Cesar must post a solution.

so, someone has a solution for this unsolved problem??

Funny how some people suddenly "forgot" about this topic...

People, were it a significant result, I would have understood a lot of discussions, but when it's about a problem about nothing, which moreover turned out to be a fake (at least, I don't think Upul Zarec's solution's is or was correct), I understand why nobody bothers. It is exactly as if the problem were to solve the equation $x^{10972}-24367 x^7+54=0$. Nobody can solve it (I mean, give a solution with radicals), but nobody cares about it.   ;)

Which was the problem, by the way? How did you get the problem?


MATHCOUNTS AMC AIME quadratics trigonometry logarithms algebra

The title pretty much says it all.  I like to use the Art of Problem Solving series. Does anyone know anything else I can use?  :D

I use the MATHCOUNTS School Handbooks, past competitions, and the All-Time Greatest MATHCOUNTS Problems. Most of those can be ordered through AoPS. Believe it or not, AoPS texts can really help on MATHCOUNTS.

The Intro Series are the best Mathcounts preperation books.

[quote="perfect628"][quote="mihail911"]Everyone knows AoPS texts are ideal for competition preparation, and way beyond mathcounts.
It's funny how mathcounts is considered one of the smaller, less difficult competitions around here. XD
I mean look at the math jams--as soon as it's about mathcounts, it has a max 2-3 rating--whereas AIME/USAMO can go up to 7-9.
I don't know why, but i find that funny.[/quote]

b/c comparitively MC isnt all that tough. comparitively. recall that MC is a middle school competition, while most mentioned on aops are high school ones.[/quote]
I agree, and MATHCOUNTS is mostly speed.  The problems don't usually require long, deep thinking, as they are meant to be dealt with in 1-3 minutes.

[quote="i_like_pie"]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=133189[/quote]

The majority of content in that particular thread is copyright discussion; the only useful links are here:

http://mathcounts.saab.org/mc.cgi/
http://www.yashar.biz/mathcounts/
and the obvious one
http://www.mathcounts.org/

i agree with temperal. iv perused that entire thread, and all it discusses is copyright laws. not very useful. i lke the sites temperal mentioned.
saab.org is also useful for SAT practices/verba/ectr. kinda off-topic but useful nonetheless

I basically use AOPS books. When I get bored i got to the all time greatest mathcounts book

:o you get bored with AoPS texts?

tsk, tsk, for shame.  :P

I know I'm a high schooler, but I'm going to say that it's important to do both Mathcounts and AMC/AIME problems when you're in middle school as preparation for Mathcounts. Had I done some AIME problems, I probably would have scored much better at National Mathcounts than I did. You never know whether the Mathcounts test will be extraordinarily hard, so a little AIME-level experience may be helpful. Don't discount Mathcounts problems, though, either.

Well said.

I myself use the AoPS 1 and the Intro books, but I find that reading the Intro books is more fun because

1. I have an old edition AoPS 1 so it looks less graphically appealing (i know, lame reason, but....)
2. The Intro books go into much more detail on each topic.

Mathcounts problems are simpler in mathematical knowledge required, but are sometimes less straightfoward.  B/c mc is a middle school competition, they cannot explicitely use high school math.  I have never seen a quadratic equation that can't be solved w/ guess and check, nor have I seen any trigonometry, logarithims, etc.  Theoretically, u could get to nats w/o high school math, but in reality, most ppl that make it to states already have hs math knowledge

Back in the day, preparation was done with the original AoPS Volume 1, and by doing past contests (you can buy problem books). Then I worked on Volume 2 after I finished Volume 1 twice, because you can still learn a lot of stuff the second time around. Mainly it's just about speed and accuracy; AoPS can teach you concepts to solve the problems, but mostly it's just practice on past contests until you've seen all the different types of problems they offered and can look at a problem and solve it in 30 seconds or whatever because you've already seen similar stuff before, spending a bit longer on harder problems.


I got a question concerning a page number.

Normally, page number appears automatically at the center of the bottom. Is there anyway you can say, "put a box around the page number"? Also, is there anyway I can change the page number (Like change the page number in the second page to be 1, instead of 2).

This is for USAMTS on next round (this round, I couldn't find a way to do this).

Thanks.

\setcounter{page}{1} resets the page number to 1.

If you're using the fancyhdr package, then you can print the page number (or anything else at the bottom of the page) however you like.  The \thepage command prints the page number.  For example:
[code]
\usepackage{fancyhdr}
\cfoot{\fbox{\thepage}} % prints page # in a box center on the bottom of each page
\pagestyle{fancy}
[/code]


In a three-digit number, the hundreds digit is greater than 5, the
tens digit is greater than 4 but less than 8, and the units digit is
the smallest prime number. How many three-digit numbers
satisfy all of these conditions?

The smallest prime number is 2.
There are 4 ways to choose the hundreds digit (6,7,8,9)
There are 3 ways to choose the tens digit (5,6,7)

Since our choices are independent of each other, we multiply out: 1*4*3=12.


A convex hexagon is inscribed in a circle.  Its successive sides are 2,2,7,7,11, and 11.  Compute the diameter of the circumscribed circle

You can always rearrange the sides of a cyclic polygon, because of course each side has a corresponding arc to it.

sorry but how does rearranging the sides help?

rearrange to 2,7,11,2,7,11, then you know the chord that splits the pair is the diameter.

You don't need to do even that.

[hide="Solution"]
Label the central angles as such: $2A$ subtending the side with length $2$, $2B$ subtending the side with length $7$, and $2C$ subtending the side with length $11$. We see that
\[4A+4B+4C=360 \implies A+B+C=90 \implies \cos(A+B) =\sin(C) \ \ \ \ \ \ \ \ \ \ \ \ (1) \]
Drawing the altitudes of the three non-congruent isosceles triangles, we see that $\sin A= \frac{2}{d}$, $\sin B = \frac{7}{d}$, and $\sin C = \frac{11}{d}$. Consequently, $\cos A = \frac{\sqrt{d^{2}-4}}{d}$ and $\cos B = \frac{\sqrt{d^{2}-49}}{d}$.

Now, plugging these into $(1)$,
\[\left( \frac{\sqrt{d^{2}-4}}{d}\right) \left( \frac{\sqrt{d^{2}-49}}{d}\right)-\frac{14}{d^{2}}= \frac{11}{d}\iff (d^{2}-4)(d^{2}-49) = (11d+14)^{2}\]
\[\iff 0= d^{3}-174d-308= (d-14)(d^{2}+14d+22) \]
Both solutions to the quadratic are negative; thus, we have  $\boxed{d=14}$.[/hide]

I knew why I recognize this problem immediately. I posted it before.

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=566095&search_id=824340823#566095

Locked.

P.S. This is no offense to any new users but I'm just doing job of reducing threads that post same stuffs over and over again.


geometry
circumcircle
conics
parabola
geometry open

Let $ABC$ be an equilateral triangle with circumcircle $(O)$.Draw a circle $(O_{1})$ which is tangent to both side $BC$ and arc $BC$.Construct the circles $(O_{2});(O_{3})$ similarly for the remaining sides $CA$ and $AB$.prove that $AO_{1}=BO_{2}=CO_{3}$ iff(ie.if and only if) the circles $(O_{1}),(O_{2})$ and $(O_{3})$ all have the same radius .
(Notation:$(X)$ is a circle with center at $X$)

denote with $ M_{1}$ the midpoint of BC, $ M_{2}$ of CA and $ M_{3}$ of AB; $ P_{1}: OO_{1}\cap (O)$ and $ Q_{1}$ is the projection of $ O_{1}$ on BC and equally for $ P_{2}$, $ P_{3}$ and $ Q_{2}$, $ Q_{3}$;
$ N_{1}$ the midpoint of the minor arch BC, $ N_{2}$ of CA and $ N_{3}$ of AB;
$ K_{i}$ is the simmetric of $ M_{i}$ with $ N_{i}$, $ r_{i}$ the line perpendicular to $ OM_{i}$ from $ K_{i}$ and $ H_{i}$: $ Q_{i}O_{i}\cap r_{i}$  (where $ i\equal{}1,2,3$)

it's easy to prove that the locus of the $ O_{i}$ is a part of a parabola with focus $ O$, directrix $ r_{i}$ and axis $ OM_{i}$ couse $ O_{i}H_{i}\equal{} R\minus{}O_{i}Q_{i}\equal{} R\minus{}O_{i}P_{i}\equal{} OO_{i}$ (where $ i\equal{}1,2,3$)

so if the 3 circle have the same radius $ M_{1}Q_{1}\equal{} M_{2}Q_{2}\equal{} M_{3}Q_{3}$ and for pitagora $ AO_{1}\equal{} BO_{2}\equal{} CO_{3}$
and if $ AO_{1}\equal{} BO_{2}\equal{} CO_{3}$, $ M_{1}Q_{1}\equal{} M_{2}Q_{2}\equal{} M_{3}Q_{3}$ and the 3 circle have the same radius  :maybe:


i am in baltimore sitting in a one room apartment pouring over some geometric number theory while it's christmas outside and everyone is so happy.

i am so lonely :(  .  i feel sad.

Have fun.

I have no christmas either.

I lost.

[quote="hunter34"]

I lost.[/quote]

Dope, I just lost the Game. I don't celebrate Christmas either. I have to study Chemistry and Mathematics over my holiday in preparation for the Chemistry and Mathematics Olympiad Exams in January, February and March.

Christmas is just like any other holiday really (I think). Just think of it as an extended weekend. Or just go outside and have fun!

[quote="BanishedTraitor"][quote="ProtestanT"]don't be lonely. Just knock on a neighbor's door and ask if they wanna have christmas with you.[/quote]

Let's celebrate Christmas on AoPS![/quote]

Good Idea, but one problem:

EVERYONE JUST LOST!

I get to celebrate Christmas, yay.

[url=http://www.losethegame.com/]Click Here[/url]

I lost

[quote="sexynsmartjenny"]i am in baltimore sitting in a one room apartment pouring over some geometric number theory while it's christmas outside and everyone is so happy.  [/quote]

I'll make you feel better:

I cannot relax during this break as I have multiple chapters to read and annotate in AP bio, a tesselation project due after break, and an english essay to do that's also due after break. Also, I have to prep for the AMC/AIME (hopefully my AoPS vol 2 will arrive ASAP), and if I have time, I can do the USAMTS problems. sigh...

[quote="aanjohri"]what is everyone talking about???
lost what?
srry, i am completely clueless[/quote]

Don't worry you just lost the game.

My christmas... doesnt really suck.. just kinda quiet and boring... and w/o presents

I'm sitting here doing some math... and am stumped on a stupid problemm....

I'm getting my presents later... on top of presents.. my dad is also signing me up for 2math courses :) yay

xmas is over, but i got one present that came late and isnt even officially mine, and im in the 5th grade. child abuse always comes early..

I lost.

Christmas is waaaay over, and I went to Jamaica (not to brag, of course). But there was no Internet, so no AoPS :(

oh. i stayed up until midnight for the past week b/c i had to fill out high school applications...

1=2, don't worry - it's New Year's eve that matters.

you should tell it do stop boring because it's already drilled deep enough.  :lol:

Ok Jenny, Out of 3 topics made here, 3 became spam. 3 Homeruns is a very good thing.

I lost.

[quote="ZhangPeijin"]Getting to spam now...[/quote]

Agreed.


Putnam algebra function domain calculus derivative geometry

http://books.google.com/books?id=AA-lOA1nPDcC&pg=PR8&dq=putnam+competition
On page four is question 1986 A1.  Also on page 9 is 1988 A1.  I'd really appreciate it if someone could quickly read over my solutions and check if they're correct (on the google reader the page with the solution isn't available for viewing).

Here's what I did for 1986 A1:
1) Find the critical values of $ f(x)$, which were $ x \equal{} \minus{} 1, 1$.
2) Those would be where the f would have its min/max if the domain wasn't restricted, but we have a restriction of x from [-3,-2] to [2,3], so we must find a max/min in one of those intervals.
2) I graphed the function and found that the max would be from [2,3] by looking at the function's general shape.  I proved that the function was increasing in [2,3] by taking the derivative of f and showing that it was positive for [1, infinity] (so f is also increasing in [2,3])
3) Because for all x in [2,3], 3>x, by definition of an increasing function, f(3) > f(x) for all x = [2,3], so f(3) is the maximum value, which is 18.

For 1988 A1:
1) Split the first equation into four cases, in which (x,y) = (+,+), (+,-), (-,+), (-,-).
2) Graphed the function for each case, saw that it looked like two cones, one right side up, one inverted, squished together.
3) Saw where it intersected with |y| < 1, and calculated the area to be 6.

I found these problem to be extremely easy.  If I got these right, am I correct in thinking that this is a problem that is easy even for a #1 for the Putnam or am I getting better?

Old Putnam problems are indeed about as easy as you think they are. For actual
competitions, you want be practicing off of newer contests, going back no more
than 10 years or so.

EDIT: Although hard problems are always worth doing.


Sally is given $100 and must spend all of it to buy exactly 100 animals.  The prices are as follows:  Cows-10 dollars, Sheep-5Dollars, Pigs-2 dollars, Chickens-50 cents. What is the maximum number of chickens that she could buy?

Since you seem to be asking for help and getting none :

[hide]the number of chickens must be even so that the you spend a whole number of dollars on chickens.  I just started with a big even number and worked my way up until it was obvious that you couldn't buy that many.  I got the answer on my second try, so maybe I am just a good guesser :lol:

I did start though with 100 chickens would be 50 dollars, and if I traded a chicken for a cow, then after 5 trades I'd be up to 97.50.  Not quite there, but close.  So I knew it would be an even number close to but less than 95.

Does that thinking help?
[/hide]

I think the answer's.....wait, do u HAVE to use ALL of the $\$$100?
-jorian[/hide]

Jorian buddy, you REALLY need to get better at reading the question carefully.

[quote]must spend all of it[/quote]

sorry, do you know what i did?

i looked at the problem (should've quoted it)

and i thought i hadn't seen it so now i feel REALLY STUPID!

sheep 5
pig 2
cow 10
chicken .50

45 -90

7 5's
2 10's

ok, i got really close... that might help :)

This can be solved with a classic application of Diophantine equations (well above MC level), so if you are interested in these, I suggest you look into them.

Like frost13 said
[quote]the number of chickens must be even so that the you spend a whole number of dollars on chickens[/quote]
[hide]We must buy 100 animals with 100 dollars. If we bought all chickens,
[b]10,000/50 = 200[/b]There are 200 chickens and the exact number is 100.
There are 50 pigs if we buy all pigs.
There are 20 sheeps if we buy all sheeps
There are 10 cows if we buy all cows.
Cows- 10 dollars substitute as A
Sheep- 5 dollars substitute as B
Pigs-2 dollars substitute as C
Chickens- 50 cents substitute as D
[b]A + B + C + D = 100
10A + 5B + 2C + 0.5D = 100[/b]
D has to be the greatest possible.
I hope this Helps a lot!!![/hide]

The question I have is this: Is it required that at least one of each animal is purchased?

[hide]I think the answer is 92 chickens, 4 cows, 2 sheep, and 2 pigs. My method was checking every even number below 98. So if you had 98 chickens then their cost would be 49 dollars leaving you with 51 dollars and 2 animals which is impossible. Then 96 chickens is 48 dollars leaving 52 dollars and 4 animals, again impossible. Then 94 chickens is 47 dollars leaving you with 53 dollars and 6 animals. So you could have 5 cows= 50 but you have three dollars left, so thats impossible. Then 4 cows = 40 and 2 animals left. If you use two sheep then thats 10 dollars but that adds up tp 50, too less. So you have 92 chickens to buy.[/hide]

Another Rose-Hulman question

Indianamath, please post a source to these questions if possible. There's a good reason.

Just yesterday I took the Rose-Hulman competition for the first time, for score. Had I seen your two posts eariler or had I decided to take the test today, I would've already read up on two of the hardest problem. Had a source been listed, I would've avoided it.

Anyhow, I got this problem anyways.

Sonny is correct with 92.

The multiple choice options were [for number of chickens]:

$A)94$ $B)90$ $C)86$ $D)84$ $E)None of the Above$

What I did was check 94 -- it obviously couldn't work.

I then checked 90. 45 dollars. Add in 5 cows. 95 animals used, 95 dollars. Can't work. What about add in 4 cows? 94 animals used, 85 dollars. Add in two sheep. 96 animals used, 95 dollars. Add in two pigs. 98 animals used, 99 dollars. Add in another two chickens. 100 animals, 100 dollars.

But wait! You added another two chickens. Therefore the answer is $92$

So, inother words the answer was none of the above...

[quote="indianamath"]No, you do not have to buy one of each.  I was looking for a more algebraic solution. The way you guys do it makes me unsure that you have the right answer. I think its easier to miss something if you just use arithmetic.[/quote]

Are you sure there is an algebraic solution which falls under the mathcounts level?

I'm sure my answer (none of the above) was correct. It's easy to prove that nothing greater than 94 works. It's easy to prove that odd numbers don't work.


does partial solution gives you points??? Because on USAMTS you can get 3 out of 5 without actually solving the problem (you have to get close tho...), how about USAMO???

Yes.  From the USAMO Teachers' Manual, Instructions to be read by USAMO Participants, page 14, Item 3:
"... partial credit may be given for constructive progress toward a solution."

In addition, under Item 9:  "The USAMO is graded more strictly that most school examinations, so it is typical for students to overestimate the scores they will recieve.  The decisions of the graders are final and may not be appealed."

The 2006 AIME and USAMO Teachers' Manual is going to print this week, and will appear on the AMC website in a few days.

Steve Dunbar
AMC Director

[quote="AMCDirector"]
"The USAMO is graded more strictly that most school examinations."[/quote]

Haha, I remember this statement. It pretty much wins it for "Understatement of the Year." But like he said, partial credit is definitely worth trying for, as it is essentially the deciding factor between two people who solved the same number of problems completely.

[quote="psychicalchemist"]i think the biggest understatement of the year was "150s on the AMC 12 are much rarer than 800s on the SAT, even the Math IIC"[/quote]

Ha, more like biggest understatement of the decade!

To compare, I got 800 on the math SAT in 8th and 100.5  :blush:  on the AMC 12 in 9th. (Obviously, I have improved a lot since 9th grade...)

Yeah... both are easy to make stupid mistakes on, but the AMC is much, much harder to know how to do all of the problems on to begin with. For example, I haven't gotten an 800 on the Math (although I got an 800 on Verbal :roll: ) simply because of one stupid mistake, and that was in 9th grade. I obviously knew how to do all the problems. On the AMC12, I'd be very happy to be able to get a 150 in 12th. :P

[quote="psychicalchemist"]i think the biggest understatement of the year was "150s on the AMC 12 are much rarer than 800s on the SAT, even the Math IIC"[/quote]

:lol::lol::lol: where did this come from?


A composite positive integer is a product $ab$ with $a$ and $b$ not necessarily distinct integers in $\{2,3,4,\dots\}$. Show that every composite positive integer is expressible as $xy+xz+yz+1$, with $x,y,z$ positive integers.

If $ n\equal{}a\cdot b$ with $ a,b>1$ is composite, take $ x\equal{}a\minus{}1, y\equal{}b\minus{}1, z\equal{}1$, we then get $ xy\plus{}yz\plus{}zx\plus{}1 \equal{} (a\minus{}1)(b\minus{}1)\plus{}a\minus{}1\plus{}b\minus{}1\plus{}1 \equal{} ab \equal{} n$.


(1) r1, r2, ... , r100, c1, c2, ... , c100 are distinct reals. The number ri + cj is written in position i, j of a 100 x 100 array. The product of the numbers in each column is 1. Show that the product of the numbers in each row is -1. (2) r1, r2, ... , r2n, c1, c2, ... , c2n are distinct reals. The number ri + cj is written in position i, j of a 2n x 2n array. The product of the numbers in each column is the same. Show that the product of the numbers in each row is also the same.

Classical, old Russian question :)

Hint: try to construct a polynomial...


Hi does anyone have an elementary way to completely prove Perron's Criterion which states that: for any polynomial $ P(x) \equal{} x^n \plus{} a_{n\minus{}1}x^{n\minus{}1} \plus{} a_{n\minus{}2}x^{n\minus{}2} \plus{}... \plus{} a_{1}x \plus{} a_0$ with integer coefficients and $ a_0 \neq 0$, if $ |a_{n\minus{}1}| > 1 \plus{} |a_{n\minus{}2}| \plus{} ... \plus{} |a_1| \plus{} |a_0|$ then $ P(x)$ is irreducible over $ \mathbb{Z}$.

[quote="Kono"]Hi does anyone have an elementary way to completely prove Perron's Criterion which states that: for any polynomial $ P(x) \equal{} x^n \plus{} a_{n \minus{} 1}x^{n \minus{} 1} \plus{} a_{n \minus{} 2}x^{n \minus{} 2} \plus{} ... \plus{} a_{1}x \plus{} a_0$ with integer coefficients and $ a_0 \neq 0$, if $ |a_{n \minus{} 1}| > 1 \plus{} |a_{n \minus{} 2}| \plus{} ... \plus{} |a_1| \plus{} |a_0|$ then $ P(x)$ is irreducible over $ \mathbb{Z}$.[/quote]
See Theorem 5 here:
http://web.mit.edu/yufeiz/www/intpoly.pdf


If $ x^2\minus{}3xy\plus{}2y^2\plus{}x\minus{}y\equal{}0$ and $ x^2\minus{}2xy\plus{}y^2\minus{}5x\plus{}7y\equal{}0,$ prove that $ xy\minus{}12x\plus{}15y\equal{}0.$

[hide="hint"]
Try to factor the first equation by first considering only the quadratic portion.
[/hide]
[hide="solution"]
$ x^2 \minus{} 3xy \plus{} 2y^2 \plus{} x \minus{} y \equal{} 0$
$ (x\minus{}y)(x\minus{}2y)\plus{}x\minus{}y\equal{}0$
$ (x\minus{}y)(x\minus{}2y\plus{}1)\equal{}0$
This gives us either $ x\minus{}y\equal{}0 \implies x\equal{}y$ or $ x\minus{}2y\plus{}1\equal{}0 \implies x\equal{}2y\minus{}1$.

$ x^2 \minus{} 2xy \plus{} y^2 \minus{} 5x \plus{} 7y \equal{} 0$
$ (x\minus{}y)(x\minus{}y)\minus{}5x\plus{}7y\equal{}0$
[i]Case 1[/i]First assume that x=y.
This gives us:
$ 2y\equal{}0 \implies y\equal{}0 \implies x\equal{}0$ which satisfies the condition $ xy\minus{}12x\plus{}15y\equal{}0$.

[i]Case 2[/i] Next, assume that x=2y-1.
This gives us:
$ y^2\minus{}5y\plus{}6\equal{}0 \implies y\equal{}2,3 \implies x\equal{}3,5$ (respectively). Both of these solutions (3,2) and (5,3) satisfy the condition $ xy\minus{}12x\plus{}15y\equal{}0$
[/hide]


search
number theory
relatively prime
number theory proposed

1)Find all primes p such that $ (7^{p\minus{}1}\minus{}1)/p$ is a perfect square number.
2)Find all primes p such that $ (19^{p\minus{}1}\minus{}1)/p$ is a perfect square number. :lol:

I've tryied to solve it for a long time. I hope anybody will give me a hint :) .

1) ... if 7 is not a primitive root mod p, i think factorisation in the a+b sqrt{2} ring works.

The enclosed file might be of some use to you.  :wink:

Search into the resources section, $ a )$ is Turkey NMO 2008 problem #2

I'm not sure that this is completely correct, but here we go:
b) $ p \equal{} 2$ is good, so assume $ p > 2$, then the expression factorises in the following way:
$ \frac {19^{\frac {p \minus{} 1}2} \minus{} 1}{2} \cdot\frac {19^{\frac {p \minus{} 1}2} \plus{} 1}{2} \equal{} p(\frac {x}{2})^2$
The two factors on the LHS are relatively prime, so one of them is a perfect square, the other is p times a perfect square.

1)
$ \frac {19^{\frac {p \minus{} 1}2} \plus{} 1}{2} \equal{} a^2$ which is equivalent to $ 19^{\frac {p \minus{} 1}2} \equal{} 2a^2 \minus{} 1 \equal{} (a\sqrt {2} \plus{} 1)(a\sqrt {2} \minus{} 1)$
It's not to hard to prove, that 19 is a prime in the $ a \plus{} b\sqrt2$ UF ring (assume $ 19 \equal{} (k \minus{} b\sqrt2)(c \plus{} d\sqrt2)$, from which $ kd \equal{} bc$ follows, so there are x,y,z,v integers such that $ k \equal{} xy$, $ d \equal{} zv$, $ b \equal{} xz$ and $ c \equal{} yv$. using these substitutions, we get $ vxy^2\minus{}2vxz^2\equal{}19$. This can hold if $ |vx|\equal{}19$, which means that 19 is a prime.), so $ 19^e \cdot (f \plus{} g\sqrt2) \equal{} a\sqrt2 \plus{} 1$  and $ 19^h \cdot (i \plus{} j\sqrt2) \equal{} \sqrt2a \minus{} 1$ for some integers e,f,g,h,i,j. Adding these, we get $ 19^e \cdot (f \plus{} g\sqrt2) \plus{} 19^h \cdot (i \plus{} j\sqrt2) \equal{} 2$. 2 and 19 are coprimes, so h=0 or e=0, and $ (f \plus{} g\sqrt2)(i \plus{} j\sqrt2) \equal{} 1$ as 19 is a prime in this ring, so one of the factors is a unit in the factorisation of $ 19^{\frac {p \minus{} 1}2}$. But if $ a \plus{} b\sqrt2$ is a unit, so is $ a \minus{} b\sqrt2$, so both of the factors are units, so only h=e=0 can hold.

2)
$ \frac {19^{\frac {p \minus{} 1}2} \minus{} 1}{2} \equal{} a^2$ which is equivalent to $ 19^{\frac {p \minus{} 1}2} \minus{} 1 \equal{} 2a^2$, and $ \frac {19^{\frac {p \minus{} 1}2} \plus{} 1}{2} \equal{} pa^2$, so $ p|19^{\frac {p \minus{} 1}2} \plus{} 1$, so $ 2a^2\equiv \minus{} 2 (mod p)$, so p is a prime of form $ 4k \plus{} 1$, so $ 2k \equal{} \frac {p \minus{} 1}2$.
Using this we get:
\[ 19^{\frac {p \minus{} 1}2} \minus{} 1 \equal{} 19^{2k} \minus{} 1 \equal{} 4\cdot \frac {19^k \minus{} 1}{2}\cdot{19^k \plus{} 1}{2} \equal{} 2a^2
\]
, which is equivalent to
\[ \frac {19^k \minus{} 1}{2}\cdot\frac {19^k \plus{} 1}{2} \equal{} 2(\frac {a}{2})^2
\]
The factors on the LHS are relatively prime, so one of them is a perfect square, the other is two times a perfect square.

2.1.
$ \frac12 (19^k \plus{} 1) \equal{} 2x^2$, $ \implies$ $ 19^k \plus{} 1 \equal{} (2x)^2$ $ \implies$ $ 19^k \equal{} (2x \plus{} 1)(2x \minus{} 1)$ $ \implies$ the factors are relatively prime, so one of them must be 1, no solution

2.2.
$ \frac12 (19^k \minus{} 1) \equal{} 2x^2$, $ \implies$ $ 19^k \minus{} 1 \equal{} (2x)^2$ $ \implies$ $ 19^k \equal{} (2x \plus{} i)(2x \minus{} i)$. 19 is a prime among gaussian integers, and the difference of the two factors is $ 2i$, so $ (2x \plus{} i,2x \minus{} i)|2i$. as $ (19,2i) \equal{} 1$ we get no solutions.

Sum: p=2 is the only solution.

Ps. Completed, thanks to jUaNi-s help.


blogs
number theory unsolved
number theory

let $ p$ a prime number then prove that :

$ q / x^{p\minus{}1} \plus{} x^{p\minus{}2} \plus{} .... \plus{} x \plus{} 1 \implies q \equal{} 1[p]$

thanks

What if $ q \equal{} p, x \equal{} 1$?  :)

See [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=155461]an old blog post[/url] for the correct generalization.

http://www.mathlinks.ro/viewtopic.php?t=196186

Stop multiposting!!!


Artificial Intelligence (AI)
Will it ever be created? Do you fear it?
Can it outsmart us?

>>Will it ever be created? Do you fear it?
>>Can it outsmart us?

Why use the future ?? AI exists, now.

The Chess champion is a computer ....

Really? I know that the match between Kasparov and the computer (I don't know it's name ended in a draw).
AI is about the computer becoming intelligent and not the perfect machine for following orders.
That chess computer does not come with new ideas it just analyises all the possibilities. I don't know the system in which the match was organized, but if both players played alternatively the white pieces, I heard that it was proved that the white has a winning strategy. This means that if the computer and the man each won their white starting matches, then the man made no mistake which is smth realy great. The computer cannot miss unless it uses (WIN95 :D) so if the strategy exists then the draw is all we can hope for.
Back to AI. It will exists when it will prove creative without testing all the possibilities for an action and decide for the less risky. Also it will exist when it will come with ideas and have a self consciousness or awareness. (Like when EVE ate the apple from the snake in the Bible)

[quote="Lagrangia"]Just watch Terminator1,2 and 3! :D[/quote]
rather bad example Lagrangia. Arnold is just takin' orders from the future. He is not quite an AI example :D

I would choose the Matrix series for this matter. Unfortunately, the end is very near. One way or another.

about AI...
i don't think that it will be create soon
it's a very very very complicated thing
i know because i made a (very little) game with something like AI and a friend of mine made a chess (which is preety good)... and i know that it's a lot and very hard to work on it
if someone want's i'll put some tutorials(or smth like this) online.
anyway i was curious sometime how can we made a little AI... and when i've heard... i said: it doesn't seems to be so hard... but it's a lot of work... and it's very hard to
imagine that if u want to make a robot which has AI... like a human... imagine how many things has to know it... IT HAS TO THINK by hisself, it has to feel, etc
anyway try kuba.port5.com/Artificial Neural Networks.rar .... i hope it works... maybe it will help u with smth
if not... sorry... i don't know so much english to say all what i know :(
cheeers :D  :D
Kuba

by the way
that's my opinion... it doesn't have to be like this  :D  :D

I don't know if that is too hard (of course it is hard if no one did it untill now).
Think about the human mind. As a child it neads to gather as much info as it can. Of course such a robot should have some basic "instincts" implemented.
I think that the machine must start by imitating the behaviour of humans around it.
Then it must learn to associate things with what we think of them.
...
The way human mind works is just too hard to grasp. Now I'm too tired and my sentences don't make any sense to me.

I've just found an article in PC Magazine (the number from October) about AI...
they made a prototype of AI... and they started to test it. :D
sooo.... now i think that (relatively) soon will be created the AI(a complete one)... this one simulate a human brain with 20 miliards neurons(!!!!)..

People have been testing AI-relatd projects for quite some time now (I saw that on the Discovery channel  :D ), but I think 'the real thing', the actual simulation of a neural net with tens or hundreds of billions of neurons will become functional many years from now...

Wow I wasn't even born when the posts from this topic was created....

I was born before the first post here, but I was less than a month old then.

Now, my beliefs:

Yes, AI has already been created. Yes, it already outsmarts us at things like Go and chess, as well as making eerily accurate predictions about whatever topic you like, if you have enough data, as in millions and millions of observations.

The DeepBlue AI uses game-tree analysis, which is fine for chess, but fails spectacularly for Go and other games because there are way too many options for a single move. Today, we use neural nets. They're much better in terms of computational requirements and accuracy, but much harder to decipher.

Finally, the last criterion amfulger gave in #4 is way, way too high, and besides, nobody really knows where self-consciousness comes from.


On the chord $\overline{AB}$ of circle $k$ with middle point $M$ there is a point $Q$ which is not equal to $A$ or $B$. Through the points $A,M$ and $Q$ runs a circle $k_1$ which intersects the circle $k$ in the points $A$ and $C$. Prove that the segments $\overline{QB}$ and $\overline{QC}$ are equal in length.

Can someone explain it: $A,M,Q$ are all on chord $AB$, so they are collinear!?

[quote="peeta"]a chord is the straight line between two distinct points on a circle.
$M$ is the mid point of the circle $k$, and not necessarily on $\overline{AB}$ but $Q$ is.

peeta[/quote]
Thanks...can you explain what the mid point of circle means?

thats the point from which all points on the circle have the same distance of the radius.
its the mid point because it is in the middle of it

This might have been confusing because the more common term is center/centre...
But I've got the problem now...nice one! :D
Maybe this will clear up some confusion...
$AB$ is a chord of a circle with center $M$.  There lies a point $Q$ on $AB$.  The circumcircle of triangle $AMQ$ intersects the circle again at $C$.  Prove that $QC=QB$.  Just draw a picture and it'll be much easier to understand.
As for how to solve it...really don't know...i'm not good at these, this will take some thinking...

Here we go: finally a graph for clarification. Hope this helps.... :idea:

Hi all!  :roll:

Since [tex]\angle QMC=\angle QAC=\angle BAC=\frac{1}{2}\angle BMC[/tex] and [tex]\angle BMC=\angle QMC+\angle BMQ[/tex] we get [tex]\angle BMQ=\angle QMC[/tex]. Next, we have [tex]MB=MC[/tex] and the common side [tex]MQ[/tex], whence the triangles [tex]QBM[/tex] and [tex]MCQ[/tex] are congruent. Hence [tex]QC=QB[/tex], what was to proof.

Hanno

hey hanno
welcome! another german on this forum :D
but dont tell me that u havent seen the proof somewhere before  ;)
peeta

Hey!

"I haven't seen the proof before" :P :P

Hanno

One question: while it's obvious that $\angle QAC = \angle BAC$, how did you determine that $\angle QMC=\frac{1}{2}\angle BMC=\angle QAC = \angle BAC$?

well, the angle "from" the center of a circle over the same chord (length) as an angle "from" the circle itself is double as long as the other one.
ok i confess that sentence is a little confusing.
but here is a picture, and so u believe me if u dont know this thereom ill prove it also  :lol:

let $M$ be the center of the circle, and $AC$ a diameter. also let $MD\perp AB$.
now we know thanks to thales that $\measuredangle ABC$ is also a right angle. so $MD \parallel CB$. so $\measuredangle MBC = \measuredangle BCM$ since $MB=MC=r$ and $\measuredangle MBC = \measuredangle DMA$, $\measuredangle BCM = \measuredangle BMD$ because of the parallels. so $\measuredangle BMA = 2\measuredangle BCA$ and this is also true for every $\measuredangle BEA$ since they are over the same chord


function
algebra
functional equation
IMO Shortlist

Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$, satisfying \[
f(xy)(f(x) - f(y)) = (x-y)f(x)f(y)
\] for all $x,y$.

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]  :)

Assume $f(1)=0$. Take $y=1$. We get $f^2(x)=0,\ \forall x$, so $f(x)=0,\ \forall x$. This is a solution, so we can take it out of the way: assume $f(1)\ne 0$.

$y=1\Rightarrow f(x)[f(x)-f(1)]=(x-1)f(x)f(1)$. We either have $f(x)=0$ or $f(x)=f(1)x$, so for every $x$, $f(x)\in\{0,f(1)x\}$. In particular, $f(0)=0$.

Assume $f(y)=0$. We get $f(x)f(xy)=0,\ \forall x$. This means that $f(a),f(b)\ne 0\Rightarrow f(\frac ab)\ne 0\ (*)$ ($\frac ab$ is defined because $f(b)\ne 0\Rightarrow b\ne 0$). Assume now that $x\ne y$ and $f(x),f(y)\ne 0$. We get $f(x)=f(1)x,\ f(y)=f(1)y$, and after replacing everything we get $f(xy)=f(1)xy\ne 0$, so $x\ne y,\ f(x),f(y)\ne 0\Rightarrow f(xy)\ne 0\ (**)$. Assume now $f(x)\ne 0$. From $(*)$ we get $f(\frac 1x)\ne 0$, and after applying $(*)$ again to $a=x,b=\frac 1x$ we get $f(x^2)\ne 0\ (***)$. We can now see that $(**),(***)$ combine to $f(x),f(y)\ne 0\Rightarrow f(xy)\ne 0\ (\#)$.

Let $G=\{x\in\mathbb R|f(x)\ne 0\}$. $(*)$ and $(\#)$ simply say that $(G,\ \cdot)$ is a subgroup of $(\mathbb R^{*},\ \cdot)$.

Conversely, let $G$ be a subgroup of the multiplicative group $\mathbb R^*$. Take $f(x)=\{\begin{array}{c}f(1)x,\ x\in G\\0,\ x\not \in G\end{array}$. It's easy to check the condition $f(xy)[f(x)-f(y)]=(x-y)f(x)f(y)$.

[quote="Pascal96"]Please tell me if there is anything wrong with my solution:
Put y=1. If $f(1)$ = 0, f is identically 0. Otherwise f(x) = f(1)x = kx for any real k.[/quote]
Yes, just look at the previous post which gives a lot of other solutions.

And since you just give the result and not the content of your own proof, we can not show you where is your error. The only thing we can say is that your proof is wrong.

I have the same solution as pascal, but i'll post it so you can point out what's wrong:
let our assertion be $P(x,y)$.
$P(x,1): f(x)(f(x)-f(1))=(x-1)f(x)f(1)$
Therefore, either $f(x)=0$ or $f(x)-f(1)=(x-1)f(1)$
Therefore, $f(x)=0$ is a solution
Now assume $f(x) \not=0$
then $f(x)-f(1)=xf(1)-f(1) \implies f(x)=xf(1)$
Plugging in $x=1$, we see $f(1)=f(1)$, so we have no further restrictions on $f(1)$
Therefore, for all $c \in \mathbb{R}$, there is an $f(x)=cx$.
Furthermore, $f(x)=0$ is merely $f(x)=cx, c=0$, so all solutions are in the form $f(x)=cx$ for all $c \in \mathbb{R} \blacksquare$

You said this is wrong; why?

It's wrong because you only have $f(x) = xf(1)$ or $f(x) = 0$ for each $x \in \mathbb{R}$. You have to do a little more work to restrict which $x$ you have fall under the first case and which fall under the second.

Oh, tired me might realize; Is it because this function is not continuous?

First, suppose $f(0)\neq 0$. Then $f(0)\cdot [f(x)-f(0)] = x\cdot f(x)\cdot f(0)$. Taking $x=1$ we get $f(1)-f(0)=f(1)$, contradiction. Thus $f(0)=0$.

Now, suppose there was a value of $x$ such that $f(x)\neq 0$. Then $f(x)\cdot [f(x)-f(1)] = (x-1)\cdot f(x)\cdot f(1)$, or $f(x)=x\cdot f(1)$. Clearly $f(1)\neq 0$, or this gives a contradiction. Moreover, if $f$ is a solution, then $c\cdot f$ is a solution for any constant $c$. Then we either have $f(1)=0$, in which case $f\equiv 0$, or $f(1)=1$, in which case $f(x)$ is either $0$ or $x$.

Let $S$ be the set of values for which $f(x)=x$. If $S$ is nonempty, then $1\in S$. Moreover note that $0\not\in S$. Now, suppose $|S|>1$. We claim that $S$ is a subgroup of $\mathbb{R}^*$. To do this, we need to show that $S$ is closed under multiplication, and that $x\in S\rightarrow 1/x\in S$.

Suppose $x\in S$ but $1/x\not\in S$. Clearly $x\neq -1$. Then $f(1/x)=0$. It follows that $f(1)\cdot [f(x)-f(1/x)] = 0$. But $f(x)=x$ and $f(1/x)=0$, so $f(1)\cdot x = 0$, implying $x=0$, contradiction as $0\not\in S$. Thus for every $x\in S$, we have $1/x\in S$.

Now we need to check for closure under multiplication. First, if $x,y$ are distinct and in $S$, then $f(xy)\cdot (x-y) = (x-y)\cdot x\cdot y$, so $f(xy)=xy$. It remains to check that for $x\in S$, $f(x^2)=x^2$. This is clear if $x=1$. But note that $f(x)\cdot [f(x^2)-f(1/x)] = (x^2-1/x)\cdot f(x^2)\cdot f(1/x)$. If $f(x^2)=0$, then it follows that $f(x)\cdot (0-1/x)=0$, or $f(x)=0$, contradiction! Thus $f(x^2)=x^2$. It follows that $S$ is also closed under multiplication.

Thus, the solutions are of the form $\boxed{f(x)=f(1)\cdot x \mid x\in S; f(x) = 0 \mid  x\not\in S}$ when $S$ is a subgroup of $\mathbb{R}^*$. It is easy to check that these solutions work.

We claim the solutions are $f(x)=kx$ for $x\in G$ and $f(x)=0$ for $x\not\in G$ where $G$ is any multiplicative subgroup of $\mathbb{R}^*$. It is easy to check that this works.

Let $P(x,y)$ be the FE and suppose $f$ satisfies it. Note that
\[P(x,1)\implies f(x)(f(x)-f(1))=(x-1)f(x)f(1),\]
or that $f(x)^2=xf(x)f(1)$. Thus, $f(x)=xf(1)$ if $f(x)\ne 0$. If $f(1)=0$, then we get $f\equiv 0$, so suppose $f(1)\ne 0$, and since we can scale $f$, WLOG suppose $f(1)=1$. Let $G=\{x:f(x)\ne 0\}$.

Note that $1\in G$. Also, if $x\in G$, then
\[P(x,1/x)\implies x-f(1/x)=(x-1/x)xf(1/x)\implies f(1/x)=1/x,\]
so $1/x\in G$. Furthermore, if $x\ne y\in G$, then
\[P(x,y)\implies f(xy)(x-y)=(x-y)xy\implies xy\in G.\]
It simply remains to be shown that if $x\in G$, then $x^2\in G$. Note that
\[P(x^2,1/x)\implies x(f(x^2)-1/x)=(x^2-1/x)f(x^2)/x\implies f(x^2)=x^2,\]
so $x^2\in G$. This shows that $G\le\mathbb{R}^*$, so we're done.

Wait, how would you write such a solution in a real olympiad? After you get the relatively trivial $f(x)=xf(1)$ or $f(x)=0$ for any $x$ real, and some relations like if $f(x)$ is a solution, then so is $g(x)=cf(x)=>$Let $f(1)=1$, $P(t,\frac{1}t) => f(\frac{1}{t})=\frac{1}{t}$ if $f(t)=t$, if $f(x)=x$ and $f(y)=y$ $=>$ $f(xy)=xy$... now how do you construct a "multiplicative sub-group" from this, and more importantly how would you write down such a function anyways?

I think all the above solutions explain this itself. Basically all you have to do is to show that there is a set of real numbers which satisfy the following relations :
(1) Show that the sub-group contains the identity element i.e $1$\\
(2) Now if $x \in G \implies \frac{1}{x} \in G$ which implies the inverse element exists\\
(3) For $x,y \in G \implies x \cdot y \in G$ which is the obvious definition of a group\\
Showing all this gives $G$ is a multiplicative sub-group of $\mathbb{R}^*$

Now if we partition $\mathbb{R}^*$ into two mutually disjoint subsets $\{G\}$ and $\{\mathbb{R}^* - G\}$, then it is fairly clear to see that all such all such solutions work. Actually it is similar to the case of a point-wise trap!

I claim that $f(x)=cx$ when $x\in G$ and $f(x)=0$ otherwise is always a solution, where $G$ is a multiplicative group over $\mathbb{R}\setminus\{0\}$, and $c$ is an arbitrary constant. It is easy to verify that such solutions work.

To show that all solutions are of this form, denote the assertion $P(x,y)$. We have that $P(1,0)$ yields $f(1)f(0)=(f(1)-f(0))f(0)$. So, either $f(0)=0$, or $f(1)-f(0)=f(1)\implies f(0)=0$. So, $f(0)=0$. Now, $P(1,x)$ yields $f(x)(f(x)-f(1))=(x-1)f(x)f(1)\implies f(x)(f(x)-xf(1))=0$. So, either $f(x)=0$ or $f(x)=xf(1)$. Assume $f(1)\neq 0$, or else $f(x)=0$. If we have nonzero $x\neq y$ such that $f(x)=xf(1),f(y)=yf(1)$, then $P(x,y)$ gives that $f(xy)\neq 0\implies f(xy)=xyf(1)$, for $x\neq y$. Similarly, we get that if $f(x)=xf(1),f(xy)=xyf(1)$, then $f(y)=yf(1)$. Denoting the set $G$ to be all $x$ such that $f(x)=xf(1)$, we have $x,y\in G\implies xy\in G$ for $x\neq y$, $x,y\in G\implies \frac{x}{y}\in G$, and $1,x\in G\implies x,\frac{1}{x}\in G\implies x^2\in G$. So, $G$ is a multiplicative group, as desired.

[quote=william122]I claim that $f(x)=cx$ when $x\in G$ and $f(x)=0$ otherwise is always a solution, where $G$ is a multiplicative group over $\mathbb{R}\setminus\{0\}$, and $c$ is an arbitrary constant. It is easy to verify that such solutions work.

To show that all solutions are of this form, denote the assertion $P(x,y)$. We have that $P(1,0)$ yields $f(1)f(0)=(f(1)-f(0))f(0)$. So, either $f(0)=0$, or $f(1)-f(0)=f(1)\implies f(0)=0$. So, $f(0)=0$. Now, $P(1,x)$ yields $f(x)(f(x)-f(1))=(x-1)f(x)f(1)\implies f(x)(f(x)-xf(1))=0$. So, either $f(x)=0$ or $f(x)=xf(1)$. If we have nonzero $x\neq y$ such that $f(x)=xf(1),f(y)=yf(1)$, then $P(x,y)$ gives that $f(xy)\neq 0\implies f(xy)=xyf(1)$. So, if we define $G$ to be the set of all $x$ such that $f(x)=x$, we get that $G$ is a multiplicative group, as desired.[/quote]

technically, all you have shown is that $x,y\in G$ when $x\ne y$ implies $xy\in G$, so you still have to show $x^2\in G$, but this is not hard. I just think you might lose a point for that since it is a nontrivial required step.

Setting $y=1$, we have the following:
$$f(x)^2=xf(x)f(1)$$
If $f(1)=0$, then we have that $f(x)=0$ for all $x \in\mathbb{R}$, in the case that it isn't equal to $0$, we have that $f(x)=f(1)x$

Let's set $f(1)=c\neq 0$ and let's define the set $G=\{ y \mid f(y) \neq 0\}$
So we want to prove the following:
$$ f(x) = \begin{cases}  cx & x\in  G \\0 & x \notin G \end{cases} $$

If this holds, then $G$ must be closed under multiplication and under division
So now let's see some characteristics of $G$
1) Obviously $1 \in G$
2)If $x \in G$ and $y \notin G$, we see by the functional equation that $f(xy)f(x)=0$, since $f(x) \neq 0$, we have that $f(xy) = 0$, that means that $xy \notin G$
3)If $x,y \in G$, we have that $\frac{x}{y} \in G$, because if this didn't hold, then by the second condition we have that $y \frac{x}{y} = x \notin G$
4)From all of this we have that if $x,y \in G$, we have that $x^{-1} \in G$, and from this we have that $xy=\frac{y}{x^{-1}}\in G$
We have seen that $G$ is closed under multiplication and division, which contains $1$.
So from this we have that the solution for this functional equation:
$$ f(x) = \begin{cases} cx & x\in  G \\ 0 & x \notin G \end{cases} $$
Where $G \subseteq \mathbb{R}$ . . . :D

Let $P(x,y)$ denote the assertion, and let $f(1)=k.$\\\\ $\textbf{Claim: }$ $f(x)\in\{0,kx\}$ for all $x\in\mathbb{R}.$\\\\ $\textbf{Proof: }$ Assume $f(x)\ne 0.$ Then, $P(x,1)$ gives $$f(x)(f(x)-f(1))=(x-1)f(x)f(1)\implies f(x)-f(1)=(x-1)f(1).$$ This implies that $f(x)=kx,$ as desired. $\blacksquare$\\\\ Let $S=\{x| f(x)=kx\}.$\\\\ $\textbf{Claim: }$ If $a\in S,$ then $\frac{1}{a}\in S.$\\\\ $\textbf{Proof: }$Let $x\in S.$ The assertion $P(x,1/x)$ simplifies to $$k(kx-f(1/x))=(x-1/x)kxf(1/x)$$$$\implies f(1/x)=k/x,$$ which implies the desired conclusion.\\\\ $\textbf{Claim: }$ If $a,b\in S$ such that $a\ne b,$ then $ab\in S.$\\\\ $\textbf{Proof: }$ Suppose $f(x)=kx$ and $f(y)=ky$ for some $x,y\in\mathbb{R}.$ From $P(x,y),$ we have $$f(xy)(kx-ky)=(x-y)(kx)(ky)\implies f(xy)=kxy,$$ as desired. $\blacksquare$\\\\ It is easy to check that any function associated with a set $S$ satisfying the two conditions described above satisfies the equation.

Any working function $f$ is a particular $f_T$ where $T$ is a multiplicative group contained in $\mathbb{R}$ and
\[f_T(x)=\begin{cases}
cx&x\in T\\
0&x\not\in T
\end{cases}\]
for some nonzero constant $c$. Work with $f_T(x)/c = f_T'$ instead for convenience.
We check that such $f_T$ work in three steps; if $x\in T,y\in T$, then the condition rewrites as $xy(x-y)=(x-y)xy$. If $x\not\in T,y\not\in T$, then the condition rewrites as $f_T'(xy)\cdot0=(x-y)\cdot0$. If $x\in T,y\not\in T$, then the condition rewrites as $f_T'(xy)f_T'(x)=(x-y)\cdot f_T'(x)\cdot0$, so it is sufficient to show that $xy\not\in T$. But this follows from the observation that otherwise we would have $y=x^{-1}xy\in T$.

Take $y=0,x=1$ to see that $f(0)f(1)-f(0)^2=f(1)f(0)\implies f(0)=0.$ Observe that if $f(x)=f(y)$, then $0=(x-y)f(x)f(y)$ implies at least one of $f(x)$ and $f(y)$ is zero, so both are. That is, the only $x$ for which $f^{-1}(x)$ could contain more than one element is $x=0$. Now, take $y\in f^{-1}(0)$ and observe that $f(xy)f(x)=0$, so either $f(x)=0$ or $f(xy)=0$ for each $x$.

We now analyze the structure of $f^{-1}(x)=S$. We have observed that for each $y\in S$ and $x\in\mathbb{R}$, either $x$ or $xy$ is in $S$. Now, note that if there existed $x,y$ such that $xy\in S$ and $x,y\not\in S$, we would get a contradiction by substituting $x,y$ in. Now, observe that if $1\in S$, then any $x$ would be in $S$, so $f\equiv0$ or $1\not\in S$. Now, note that for any $y\not\in S$, $y^{-1}\not\in S$ because otherwise one of $1$ and $y$ would be in $S$. Hence, $T=\mathbb{R}\setminus S$ is a multiplicative subgroup of $\mathbb{R}$. Now, let $f(1)=c$ and define $g(x) = f(x)/c$ which satisfies the same properties as $f$ except $g(1)=1$. Take $x\in T$ and observe that the given information applied to $x,1$ yields $g(x)-f(1)=x-1$, so $x=g(x)$, giving us the claimed $f$.

Let $P(x, y)$ denote the assertion. Note that $P(x, 1)$ yields\[f(x)(f(x) - f(1)) = (x-1)f(x)f(1)\]and if we let $f(1) = c$ for some real constant $c$, then we get\[f(x)(f(x) - c) = c(x-1)f(x)\]which in fact, nicely rearranges to $f(x)(f(x) - cx) = 0 \implies f(x) \in \{0, cx\}$. First, note that $f(x) = 0$ is a solution; for the rest of the proof, we will assume $c \neq 0$. Clearly $f(0) = 0$. Let $S$ denote the set of all nonzero reals $r$ for which $f(r) = cr$. By definition, $1 \in S$. Furthermore, from $P(x, \tfrac{1}{x})$, we can see that\[c\left(f(x) - f\left(\frac 1x\right)\right) = \left(x - \frac 1x\right)f(x)f\left(\frac 1x\right)\]so if $x \in S \iff f(x) = cx$, then upon manipulation we get\[cx - f\left(\frac 1x\right) = x\left(x - \frac 1x\right)f\left(\frac 1x\right) \implies cx = x^2f\left(\frac 1x\right) \implies f\left(\frac 1x\right) = \frac{c}{x}\]so $\tfrac1x \in S$ as well. Furthermore, if we have $x \neq y$ both $\in S$, then from $P(x, y)$, we get that\[f(xy) = \frac{(x - y)c^2xy}{c(x - y)} = cxy\]hence $xy \in S$. Furthermore, if $x \in S$, we can use $P(x^2, \tfrac{1}{x})$ to deduce that\[cx\left(f(x^2) - \frac cx\right) = \left(x^2 - \frac 1x\right)f(x^2)\left(\frac cx\right)\]which upon solving for $f(x^2)$, yields $f(x^2) = cx^2$, so $x^2 \in S$ as well. Thus, the set $S$ is completely closed under multiplication, and division as well. It turns out that this is actually enough: all functions\[\boxed{f(x) =
\begin{cases}
 0 & \text{     if     } x \not\in S\\
 cx & \text{     if     } x \in S
\end{cases}}\]where $S$ is a set containing $1$ closed under both multiplication and division, work. This can be case-checked to work for all $(x, y) \in \mathbb{R}^2$, so we are done. $\blacksquare$

[b]ISL 01A4.[/b] Find all functions $f: \mathbb{R}\mapsto\mathbb{R}$ such that \[\forall (x, y)\in\mathbb{R}^2: f(xy)(f(x)-f(y))=(x-y)f(x)f(y)\stackrel{\text{def}}{:} P(x, y).\]

[b]Solution.[/b] $P(x, 0): f(0)(f(x)-f(0))=xf(x)f(0)$. This motivates us to divide into two cases as follows:

[i]Case 1.[/i] $f(0)\ne 0$
Then $(1-x)f(x)=f(0)$.
Plugging in this to the original equation we get $x+y=2 \forall (x, y)\in\mathbb{R}^2$, which is clearly a contradiction.

[i]Case 2.[/i] $f(0)=0$
If $f=0\forall x\in\mathbb{R}$ then it obviously satisfy $P$.
It suffices to consider the case when $\exists c\ne 0: f(c)=0$.
$P(x, 1): (x-1)f(x)f(1)=f(x)(f(x)-f(1))$. For $x\ne 0$, $f(x)=xf(1)$, which obviously satisfy $P$ again.

In conclusion, $f\in\{f(x)=cx\forall x\in\mathbb{R}: c\in\mathbb{R}\}$.

@above, once you get $f(xz)f(x)=0$, how do you get $f(x)=0\forall x$?

Let $P(x,y)$ be the given assertion.
$P(x,1)\Rightarrow f(x)^2=xf(x)f(1)\Rightarrow  f(x)=\begin{cases}xf(1)&\text{if }x\in S\\0&\text{if }x\notin S\end{cases}$
where $S\subseteq\mathbb R\setminus\{0\}$.
WLOG $f(1)=1$ (for ease).

[b]Case 1:[/b] $x,y\in S$
$P(x,y)\Leftrightarrow f(xy)=xy$, so $xy\in S$
[b]Case 2:[/b] $x\in S,y\notin S$
$P(x,y)\Leftrightarrow xf(xy)=0$
and since $x\ne0$, we must have $f(xy)=0$.
[b]Case 3:[/b] $x\notin S,y\in S$
Same as [b]Case 2[/b] with $x,y$ switched
[b]Case 4:[/b] $x\notin S,y\notin S$
$P(x,y)\Leftrightarrow0=0$

So the entire solution set is:
$$\boxed{f(x)=\begin{cases}cx&\text{if }x\in S\\0&\text{if }x\notin S\end{cases}}$$
where $S\subseteq\mathbb R\setminus\{0\}$ and $xy\in S\Leftrightarrow x,y\in S$ and $c$ is any real constant. This works by the casework described above.

[hide=Solution]
Substituting $y=1$, we get $$f(x)^2-f(x)f(1)=(x-1)f(x)f(1),$$ so $$f(x)^2=xf(x)f(1).$$ Therefore, either $f(x)=0$ or $f(x)=xf(1)$ for each $x$. If $f(1)=0$, then $f(x)=0$ for all $x$. Now, suppose that $f(1)\neq0$. Let $S$ be the set of values such that $f(x)\neq0$. Then, we claim that $S$ is a group with multiplication. If $a\neq b$ and $a,b\in S$, then substituting $x=a$ and $y=b$, we get
\begin{align*}
f(ab)(f(a)-f(b))&=(a-b)f(a)f(b)\\
f(ab)f(1)(a-b)&=(a-b)f(1)^2ab\\
f(ab)&=f(a)ab.\end{align*}
Therefore, $ab\in S$. Now, we will show that if $a\neq\frac1a$, and $a\in S$ then $\frac1a\in S$. Substituting $x=a$ and $y=\frac1a$, we get
\begin{align*}
f(1)\left(f(a)-f\left(\frac1a\right)\right)&=\left(a-\frac1a\right)f(a)f\left(\frac1a\right)\\
f(1)^2a-f(1)f\left(\frac1a\right)&=\left(a-\frac1a\right)f(1)af\left(\frac1a\right)\\
f(1)a-f\left(\frac1a\right)&=\left(a-\frac1a\right)af\left(\frac1a\right)\\
f\left(\frac1a\right)&=\frac{f(1)}a.
\end{align*}

Therefore, if $a\in S$ then $\frac1a\in S$. Now, we claim that $a^2\in S$. Substituting $x=a^2$ and $y=\frac1a$, we get
\begin{align*}
f(a)\left(f(a^2)-f\left(\frac1a\right)\right)&=\left(a^2-\frac1a\right)f(a^2)f\left(\frac1a\right)\\
f(1)a\left(f(a^2)-\frac{f(1)}a\right)&=\left(a^2-\frac1a\right)\frac{f(1)}af(a^2)\\
a^2\left(f(a^2)-\frac{f(1)}a\right)&=\left(a^2-\frac1a\right)f(a^2)\\
-af(1)&=-\frac1af(a^2)\\
f(a^2)&=a^2f(1).\end{align*}
Therefore, $a^2\in S$, so $S$ is closed under multiplication. Therefore, $S$ is a group.

Now, we will show that this solution works.

Case 1: $x,y\in S$
Then, $xy\in S$, so the equation is $f(1)^2xy(x-y)=(x-y)f(1)^2xy$, which is true.

Case 2: $x\in S$, $y\not\in S$ or $x\not\in S$, $y\in S$
Since swapping $x$ and $y$ multiplies both sides by negative $1$, assume without loss of generality that $x\in S$ and $y\not\in S$. The equation becomes $f(xy)f(x)=0$. We claim that $f(xy)=0$. Suppose that $xy\in S$. Then, $xy\cdot\frac1x\in S$, which is a contradiction. Therefore, $f(xy)=0$, so the equation is true in this case.

Case 3: $x,y\not\in S$.
Then, the equation becomes $0=0$.

Therefore, the only solutions are $\boxed{f(x)=\begin{cases}0&x\not\in S\\cx&x\in S\end{cases}}$, where $S$ is a group with multiplication.

Let $P(x,y)$ be the assertion.
$P(x,1) \implies f(x)^2=xf(x)f(1).$
Assume $f(1)\neq 0,$ and let $S$ be the set of all real $x$, s.t. $f(x)=xf(1).$ We then have the following (possible) solutions:
\[\boxed{f(x) =
\begin{cases}
xf(1) & \text{ if } x \in S\\
0 & \text{otherwise},
\end{cases}}\]
where $f(1)$ is any real constant not $0,$ it's easy to check both of these fit.
Case 1: $1\in S$ since $f(1)\neq 0.$
Case 2: $x\in S, y\notin S,$ then $P(x,y) \implies f(x)f(xy)=0,$ so $xy \notin S.$
Case 3: $x,y \in S$ then $\frac{x}{y} \in S.$ And by Case 2, $\frac{1}{x}, \frac{y}{x} \in S.$
To conclude, $S$ is a set containing $1$ but not $0$ and is closed under multiplication and division.

Let $P(x,y)$ denote the assertion. $P(x,1)$ gives $f(x)(f(x)-f(1))=(x-1)f(x)f(1).$ For each $x,$ we have $f(x)=0$ or $f(x)=xf(1).$ Clearly, if $f(1)=0$ then $f(x)^2=0$ which implies $f(x)=0$ for all $x$ which is a valid function.

Let $\mathbb{S}$ be the set of $x$ such that $f(x)\neq 0.$ Suppose $\mathbb{S}$ is nonempty. We have $1\in \mathbb{S}$ but $0\notin \mathbb{S}.$ However, $\mathbb{S}$ is closed under multiplication because if $f(x)=xf(1)$ and $f(y)=yf(1)$ then $f(xy)f(1)(x-y)=(x-y)xy f(1)^2$ so if $x\neq y$ then $f(xy)=f(1)xy.$

Also, $\mathbb{S}$ is closed under division because if $x\in \mathbb{S}$ then $P(x,\frac{1}{x})$ gives \[f(1)(xf(1)-f(y))=(x-y)xf(1)f(y)\]but $f(1)\neq 0$ so $(x-y)xf(y)=xf(1)-f(y).$ If $f(y)=0$ then $xf(1)=0$ which is a contradiction. Thus, $\frac{1}{x}\in \mathbb{S}.$ Therefore, $\frac{y}{x}\in \mathbb{S}$ if $y\in \mathbb{S}$ which implies that $\mathbb{S}$ is closed under division.

Note that all functions satisfying this also works: $f(x)=0$ unless $x\in \mathbb{S}$ in which case $f(x)=kx$ where $\mathbb{S}$ has the property that two of the following properties implies the third: $a\in \mathbb{S}, b\in \mathbb{S},ab\in \mathbb{S}$ and $1\in \mathbb{S}$ and $0\notin \mathbb{S}.$

The answer is
$$f(x)=\begin{cases}cx& x \in S\\0 & x \not \in S,\end{cases}$$
where $S$ is a subset of $\mathbb{R}$ closed under multiplication and division. A case check shows that these all work.

First, $P(x,1)$ yields
$$f(x)(f(x)-f(1))=f(x)f(1)(x-1) \implies f(x)(f(x)-xf(1))=0.$$
Letting $x=0$ yields $f(0)=0$, henceforth all variables are nonzero unless otherwise specified. Further, if $f(1)=0$, then $f \equiv 0$, which is described by taking $S=\emptyset$. Henceforth suppose $f(1) \neq 0$, and note that by scaling we may assume that $f(1)=1$ (so $1 \in S$), so $f(x) \in \{0,x\}$ for all $x$.
Suppose $x,y \in S$, where $x \neq y$. Then $P(x,y)$ implies $f(xy)=f(x)f(y)=xy$, so $xy \in S$ and $S$ is closed under multiplication of [b]distinct[/b] elements.
Likewise, if $x,xy \in S$, $P(x,y)$ implies (as variables are nonzero)
$$xy(x-f(y))=(x-y)xf(y) \iff xy-yf(y)=xf(y)-yf(y) \iff f(y)=y,$$
hence $y=\tfrac{xy}{y} \in S$ as well, so $S$ is (unconditionally) closed under division.
We now show that $S$ is also closed under multiplication of non-distinct elements, which is a fancy way of saying that $x \in S \implies x^2 \in S$. Indeed, note that due to the properties of $S$ we discovered, including the fact that $1 \in S$, we have $x \in S \implies \tfrac{1}{x} \in S \implies \tfrac{1/x}{x}=\tfrac{1}{x^2} \in S \implies x^2 \in S$, which implies that $S$ is closed under multiplication and division, as desired. $\blacksquare$

Please tell me if my solution is wrong.

[hide=Solution]The answer is $f \equiv 0$, $f \equiv cx$ for all $x \in \mathbb{R}$.
Let $P(x,y)$ be the assertion
\[P(x,y)=f(xy)(f(x)-f(y))=(x-y)f(x)f(y)\]
From $P(1,0)$ we have
\[f(0)(f(1)-f(0))=f(1)f(0)\]
Assume $f(0) \neq 0$. Then we have $f(1)-f(0)=f(1) \Rightarrow f(0)=0$. A contradiction. So, $f(0)=0$.
From $P(1,-1)$ we have
\[f(-1)(f(1)-f(-1))=2f(1)f(-1)\]
Assume $f(-1)=0$. Then from $P(x,-1)$ we have
\[f(-x)(f(x)-f(-1))=(x+1)f(x)f(-1) \Rightarrow f(-x)f(x)=0\]
So, either $f(-x)=0$ or $f(x)=0$. In both cases we have $f \equiv 0$ for all $x \in \mathbb{R}$ which is a solution.
When assuming $f(1)=0$ we get $f(-1)=0 \Rightarrow f \equiv 0$. So basically the same result.
From $P(x,1)$ we have
\[f(x)^2-f(x)f(1)=(x-1)f(x)f(1)\]
where $f \equiv 0$ works. Now assume $f(1)=0$. Then we get $f(x)^2=0 \Rightarrow f(x)=0$. So, assume $f \not\equiv 0$ for $x \neq 0$. Then we have
\[f(x)^2=xf(x)f(1)\]
$\Rightarrow$
$f(x)=cx$ where $c=f(1)$. So, $f \equiv 0$ and $f \equiv cx$ are the solutions.[/hide]

Let the set of all sets of real numbers that are closed over division containing $1$ be $S$. Let $S' = S \cup \emptyset$. Then the set of all valid functions $f$ can be characterized as the functions such that for some set $A \in S'$, we have $f(x) = cx$ for $x \in A$ and $x = 0$ otherwise.

It is trivial to see that all such functions work.

$P(x, 0)$ gives either $f(0) = 0$, or $f(x) = \frac{f(0)}{1-x}$, which doesn't work, so we can conclude $f(0) = 0$. Now consider the values which upon $f$ is nonzero. Clearly if $f(x)$ is nonzero, we have $P(x,1)$ gives $f(x)(f(x) - f(1)) = (x-1)f(x)f(1)$. If $f(1) = 0$, this forces $f(x) = 0$, which is a contradiction, otherwise we have $f(x) = f(1)x$. Now, we show that the set of values which $f$ is nonzero is closed under divison. Firstly, note that $f$ is injective when it is nonzero. For division, notice that for distinct $a,b$ (the case where $a = b$ just wants us to show $1$ is in the set, which has already been shown), we have $f(a)(f(b) - f(\frac ab)) = (b- \frac ab)f(a)f(b)$, where the left side is clearly nonzero. We are done.

Solved with [b]cxyerl.[/b]
Let $G$ be any multiplicative subgroup of $R^{*}$. The answer is $f(x)=Cx$ when $x\in G$ and $f(x)=0$ otherwise, where $C$ is any real. This "clearly" works.
$P(x,1)$ gives $f(x)=0$ or $f(x)=Cx$ where $C=f(1)$ for each individual $x$, so $f(0)=0$. Now let $S$ be the set of nonzero reals $x$ such that $f(x)=Cx$. Clearly $S$ is closed under multiplication and $x\in S$ iff $1/x\in S$. This means that $S$ is a multiplicative subgroup of $R^{*}$. $\blacksquare$

[b][color=pink]Claim:[/color][/b] $f$ is injective on nonzero outputs.
[i]Proof.[/i] Suppose that $f(a) = f(b) \ne 0$.
Then $P(a, b)$ implies that $a = b$. $\blacksquare$
Let $S$ be the kernel of $f$.
First suppose that $a \in S$ such $a \ne 0$. If $1 \in S$ then by $P(x, 1)$ it follows that $f(x)$ is uniformly $0$, so suppose not.
Then, by $P(x, 1)$ \[ f(x)(f(x) - f(1)) = (x - 1)f(x)f(1) \] so either $f(x) = 0$ or \[ f(x) = xf(1) \]

[b][color=pink]Claim:[/color][/b] ${\mathbb R} \setminus S$ forms a multiplicative subgroup.
[i]Proof.[/i] For $x, y \not\in S$ then \[ f(x)\left(f(y) - f\left(\frac{x}{y}\right)\right) = \left(y - \frac{x}{y}\right)f(y)f\left(\frac{x}{y}\right) \]
so either $y^2 = x$ or $f\left(\frac{x}{y}\right)$ is nonzero, but the first implies the second.
Then, since $1 \not\in S$ it follows that \[ \frac{x}{\frac{1}{y}} = xy \not\in S \] $\blacksquare$
As such, the solution set is then \[ f(x) = \begin{cases} cf(x) & x \in H \\ 0 & x \not\in H \end{cases} \] where $H < {\mathbb R}$ is a multiplicative subgroup.

This is sweet.

Let $S$ be a multiplicative subgroup of $\mathbb R$. Then $f(x) = cx$ for a real $c$ and $x \in S$ and $f(x)=0$ otherwise works; this is evident through casework.

Firstly, if $f(a) = f(b) \neq 0$, we get
\[ 0 = f(ab) (f(a)-f(b)) = f(a)f(b)(a-b) \]
which means $a=b$.

Secondly,\[ f(0)(f(0) - f(a)) = -af(0)f(a) \implies f(0)^2 = 0. \] (by setting $a=1$)

Let $S$ be the set of reals such that $\forall s \in S$ we have $f(s) \neq 0$ and suppose it is nonempty.

We verify that it is a multiplicative subgroup.
$\textbf{Identity.}$ Note that if $f(a) \neq 0$, \[ f(a)(f(a) - f(1)) = f(a) f(1) (a - 1) \] which implies $f(a) = af(1)$. As such, $f(1) \neq 0$ which means $1 \in S$.

$\textbf{Closure.}$ We have that if $a,b$ are in $S$ and are distinct, $f(a) - f(b) \neq 0$ and $(a-b) \neq 0.$ It follows that $f(ab) \neq 0$.

$\textbf{Inverse.}$ If $a \neq 1/a$ is in $S$,
\[ f(1)(f(a) - f(1/a)) = (a - 1/a) f(a) f(1/a) \]
and after supposing $f(1/a) =0 $ we derive a contradiction. If $a = 1/a$ then $a$ is its own inverse and is in $S$.

Associativity is trivial.

This readily solves the problem.

Let $K \in \mathbb{R}$ s.t. it is closed under multiplication and division then $f(x)=cx$ works for $c \in \mathbb{R}$, verification is easy.
By $P(x,0)$, we get $f(x)=0$ or $f(x)=cx$ for every value of $x$, then by some value putting we get the claimed set as solution.


Sorry if this is the wrong area and sorry for my dumbness but i really need help working out a percentage...percentages were always my sticking point!! I had my final exam today and it went really well but then i got home and my coursework score had been emailed to me and was rubbish to say the least!! For this module there is a 60/40 split in the marking with the coursework contributing 60% of the overall mark and the exam contributing 40% of the overall mark. Well i only got 34.5% for the coursework and giving that i require 40% overall from the 2 marks to pass the module can anyone tell me what percentage i need to get in the exam to get 40% overall, given the 60/40 split 60%coursework 40%exam?? If it were 50/50 then i know 45.5% in the exam would give me the 40% overall target but this 60/40 business has me a bit stumped...thanks in advance for any feedback on this   :)


hi,

i have two questions   :D :

1)prove that :
$a,b,c >0 , a+b+c=3,p\geq1 \Rightarrow (1+a^p-a)(1+b^p-b)(1+c^p-c)\geq1$

2)is there any OM online where anybody can participate without

being obliged to have a specific nationality ?

advanced thanks .

[quote="toetoe"]
1)prove that :
$a,b,c >0 , a+b+c=3,p\geq1 \Rightarrow (1+a^p-a)(1+b^p-b)(1+c^p-c)\geq1$
[/quote]
This is Zhaobin's inequality. Try the Zhaobin's theorem:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=64933
for certain $p$.


Δίνεται τρίγωνο ABC και AD η διχοτόμος του . Αν Ε ,F είναι τα ίχνη των καθέτων από το D στην AC και AB αντίστοιχα και οι ευθείες ΒΕ και CF τέμνονται στο Η και ο περιγεγραμμένος κύκλος του AFH τέμνει την ΒΕ στο Η και στο G , να δείξετε ότι το τρίγωνο με πλευρές ΒG , GE και BF είναι ορθογώνιο .

Λοιπόν βρήκα μια πάρα πολύ ωραία λύση . Ελπίζω να σας αρέσει .
Στο σχήμα μου είναι AC>AB άρα και GE>GB .
[color=red]ΛΗΜΜΑ [/color]
Έστω Κ το σημείο τομής της ΑΗ με τη ΒC τότε $AK\bot BC$ .
[color=red]ΑΠΟΔΕΙΞΗ[/color]
Χρησιμοποιούμε το θεώρημα Ceva to θεώρημα εσωτερικής διχοτόμου και μετά δείχνουμε ότι τα τρίγωνα ΒΚΑ και BFD είναι όμοια .

[color=blue]ΑΠΟΔΕΙΞΗ ΑΣΚΗΣΗΣ[/color]
Αφού από το λήμμα $AK\bot BC$  θα έχουμε ότι το τετράπλευρο AFKD είναι εγγράψιμο άρα από το θεώρημα της δύναμης σημείου θα έχουμε $BF\cdot BA=BK\cdot BD$ (1).
Όμως από το θεώρημα της δύναμης σημείου ως προς τον περιγεγραμμένο κύκλο του AFG έχουμε $BH\cdot BG=BF\cdot BA$ (2)
Από τις (1) , (2) έχουμε ότι $BH\cdot BG=BK\cdot BD$ που η τελευταία δείχνει ότι τα τρίγωνα ΒGD και BHK είναι όμοια άρα $\angle{BGD}=90$ .
Τώρα από το πυθαγόρειο θεώρημα έχουμε $GE^{2}=DE^{2}-GD^{2}$ (3)
και $BG^{2}=BD^{2}-GD^{2}$ (4)

(3)-(4) =>  $GE^{2}-BG^{2}=DE^{2}-BD^{2}=DF^{2}-BD^{2}=BF^{2}$

άρα $GE^{2}-BG^{2}=BF^{2}$  , το ζητούμενο

***Αφιερωμένη στο Νίκο Ράπανο

ωραία λύση Σιλουανέ
 :wink:

Πολύ ωραία λύση .
Η άσκηση είναι IMO Selection team 2003 .

Τη θυμάμαι την άσκηση-την είχα λύσει πέρυσι κατά τη διάρκεια της προετοιμασίας μου (ίσως να είχε ξαναπροταθεί στο advanced...)-είναι πραγματικά πολύ ωραία όπως όλες οι κινέζικες!!

 [i]Σιλουανέ σε ευχαριστώ πολύ για την αφιέρωση, με τιμά πολύ![/i]


I just noticed that there are post numbers on the right for ML skin.  Is this a new feature :?:  I have not remembered seeing this.  It makes it a lot easier to link to a post though!
:D

I think it's a new feature. I was wondering how to post user-post specifics before, but this makes it a lot easier  :) .

PReviously, you could have used the little page button next to "Posted by..."


Solve in the set of natural numbers this equation :

$ 5^n\plus{}3\equal{}2^m$

Obvious answers are:

n = 0, m = 2
n = 1, m = 3
n = 3, m = 7

A brief test shows that there are no other nice answers to this up until and including n = 15.

Substitute x = 2 we have:

$ (2x\plus{}1)^n \plus{} (2x\plus{}1) \equal{} x^m$ + x

I have no idea where to go with this after that... It seemed like a useful substitution.


On April 24th, 2006, there will be a middle/high school mathematics contest online. Registration is free, and is coming up soon. Teams of 6, calculators AND computers allowed (no internet!!!). Not all team members must be from the same school.

Google: Purple Comet Math Meet

Great, you just had to tell people... Now I won't win :(


hello for 1) do you mean this integral $ \int\frac{1}{x\sqrt{2x^2\plus{}5x\plus{}1}}\,dx$?
Sonnhard.


This inequality is cute and not too difficult :

If a, b, c > 0 such that abc <= 1, prove that

a/b + b/c + c/a >= a + b + c.

By AM-GM :

1/3(a/b + a/b + b/c) >= (a/bc)^1/3

1/bc >= a so 2a/b + b/c >= 3a  + the same by permutations

Summing 3 such inequalities gives the result.


In how many possible ways can two distinct numbers be chosen randomly from the set of first fifteen natural numbers such that the sum of the two chosen numbers is divisible by three but not by four.

We first count the number of pairs that have a sum divisible by 3, and then subtract the number of pairs with a sum divisible by 12.

Let $ x$ and $ y$ be two numbers we choose, then

$ x \plus{} y \equiv 0 \mod 3$.

Hence, either $ x \equiv y \equiv 0$, or $ x \equiv 1$ and $ y \equiv 2$.

For the first case, there are $ \binom{5}{2} \equal{} 10$ pairs and for the second there are $ 5*5 \equal{} 25$ pairs. So, there are $ 10 \plus{} 25 \equal{} 35$ pairs that have a sum divisible by 3.

Now for the pairs that have a sum divisible by 12. Notice that the only sums that can be attained that are divisible by 12 are 12 and 24. From here we can easily count the pairs. For 12: (1,11), (2,10), ..., (5,7) = 5 and for 24: (9,15),(10,14),...(11,13) = 3.

So the answer to the problems is just $ 35 \minus{} (5 \plus{} 3) \equal{} \boxed{27}$.

EDIT: Fixed.

[quote="math_explorer"]if "natural numbers" is defined as the nonnegative numbers, the definition that is slightly less used.[/quote]  Just a cultural note: in primary and secondary school I was taught that the natural numbers begin with 1 while the whole numbers begin with 0.  "The whole numbers" is a name not used by mathematicians whatsoever.  In number theory (and maybe some related areas) it seems common to take $ \mathbb{N}$ to be the positive integers, but in combinatorics and computer science it's much more common for $ \mathbb{N}$ to mean the nonnegative integers.  I like the latter choice better, for two reasons: (1) I'm a combinatorialist, and it's the industry standard (i.e., status quo bias), and (2) it makes more sense from the point of view of the English language (since "[b]n[/b]onnegative integer" is an unambiguous term).  All told, I think it would behoove people to be explicit about what they mean and so use the terms "positive integer" and "nonnegative integer."

[quote="math_explorer"][hide="Bugfix"]you subtracted two extra cases, (6,6) and (12,12), that you never considered in the first place. The answer is $ \boxed{27}$.

I think the answer is still the same if "natural numbers" is defined as the nonnegative numbers, the definition that is slightly less used. 15 and 0 are interchangeable in the mod 3 part; (0,6) is an extra pair to eliminate, but to balance that (9, 15) is one less.[/hide][/quote]

Thanks I fixed it


I'm sorry if this isn't related, but I'm trying to create a schedule for 30 teams, in which each team plays 17 teams 6 times each, and plays 12 teams 5 times each.

Is that possible?

If so, is there a way to generate them according to some sort of formula?

Thanks.

[quote="t0rajir0u"]Essentially, you want to construct a $30 \times 30$ matrix of $5$s and $6$s with exactly seventeen $6$s in any given row and column and with all diagonal entries $0$.  This seems doable to me.[/quote]
The task t0r just described is doable: there are 5s in the six spaces above and below each diagonal entry (wrapping around, if necessary).  I assumed the author wanted something more complicated, to express this as a sum of 162 game-day schedules (symmetric 0-1 permutation matrices with 0 diagonal).  Although if teams get days off, that gives appreciably more flexibility.  I can't think of an obvious mechanical way to do this, although there might be.  What do you need it for?  (I assume you aren't working for MLB  :wink:)

Also, please don't double-post.


function
limit
calculus
integration
real analysis
real analysis unsolved

Let $\ f: [1,\infty)\rightarrow\mathbb{R}$ a differentiable function such that $\ f(1)=1$ and
$\ f'(x)=\frac1{x^2+f^2(x)}$ for any $\ x\geq1$.
Prove that $\ \lim_{x\to\infty}f(x)$ exists and it is at most $\ 1+\frac{\pi}{4}$.

$0 \leq f'(x) \leq \frac{1}{x^2+1}$. Then integrate between $1$ and $\infty$. This gives the result I think.

Is there a solution without integrals? I didn't learn them yet.

Not really, but you can sneak around it.

Since $f'(x)>0$ for all $x$, $f$ is strictly increasing. Since $f(x)>1$ for all $x>1$, $f'(x)<\frac1{1+x^2}$ for all $x>1$. An increasing function either tends to a limit or diverges to $\infty$.

Now, let $g(x)=f(x)-\arctan x-1+\frac\pi4$. $g(1)=0$, and $g'(x)<0$ for all $x>1$. By the Mean Value theorem, $g(x)<0$ for all $x>1$ and $f$ is bounded above by $\lim_{x\to\infty}\arctan x+1-\frac\pi4=\frac\pi4+1$. Since $f$ is increasing, it tends to a limit which is less than or equal to $\frac\pi4+1$

Thanks  :) , it seems it's a little unnatural but is ok.


Bottleguy collects bottles. He has 400 of the 500 total different types of bottles. If he buys a 6 pack of bottles, and no two bottles in the pack are the same, what is the percent chance he will get a new bottle?

There are no 2 packs with the same bottles. Correct?

[quote="nutz_for2.718281828"][hide="solution"]Well, let's use complementary counting.The probability that no bottles are new is $\frac{400*399*398*397*396}{500^{5}}$which is approximately .31955. Therefore, the answer is $.6804405757$[/hide][/quote]

He said percent, so it's $\boxed{68\%}$

:oops_sign:  :wallbash_red: Why did I miss that?

[quote="b-flat"][quote="Astronum"][hide][b]1.[/b] [i]Assuming there aren't two packs with the same bottles:[/i]

The probability that he won't get a new bottle is:

$\frac{\binom{400}{6}}{\binom{500}{6}}= \frac{118583503}{455794085}$

Then, the probability that he will get a new bottle is:

$(1-\frac{118583503}{455794085}) \times 100 = \boxed{74\%}$[/hide]

What was my mistake, since my solution is different from [i][b]nutz_for2.718281828[/b][/i]'s solution? I confess I just start learning combinations.  :blush:[/quote]

I think that your answer is correct, for each pach containing all different bottles.  As for nutz for e's, He assumed that you could have several of the same bottles in a pack, hence the $500^{5}$ total.  This makes very little difference in the answer (it gives $74.7\%$), if you do it for 6 bottles, but he only did the answer for 5 bottles.[/quote]

Yes, and according to the problem, your solution is actually correct, Astronum.

b-flat and mysmartmouth, thank you for your clarification!  :lol:

[quote="Totally Zealous"][quote="Astronum"]
$\frac{\binom{400}{6}}{\binom{500}{6}}$[/quote]

What do the parentathesis thingys mean?[/quote]
It's a combination. You may know them as $_{400}C_{6}$.

[quote="Totally Zealous"][quote="Astronum"]
$\frac{\binom{400}{6}}{\binom{500}{6}}$[/quote]

What do the parentathesis thingys mean?[/quote]

$\binom{400}{6}$ means the number of ways of choosing 6 distinct items from a set of 400, where order does not matter.


Hmmm, I'm got as far as log5 < log(pi)/log(4).

[quote="nutz_for2.718281828"]Are you allowed to use a regular scientific calculator that can't compute logs other than base 10 and e?[/quote]
Would it matter? Using logarithm properties, you can compute logarithms of any base...

[quote="Franz Joseph"][hide]
$\pi>3>\sqrt{8}=2^{\frac{3}{2}}$
$\pi>3>\sqrt{5}=5^{\frac{1}{2}}$[/hide]

The hint should have made it pretty obvious.[/quote]

Wow, nice hint, this makes it really easy.

So,

$\pi>3>\sqrt{8}=2^{\frac{3}{2}}$
$\pi>3>\sqrt{5}=5^{\frac{1}{2}}$

Therefore

$log_{2}\pi>3/2$
$log_{5}\pi>1/2$

then it follows:

$log_{2}\pi+\log_{5}\pi> 3/2+1/2 = 2$

And we are done.


Ισως οχι και τοσο σχολικη....
Να αποδειχθει οτι αν η ορισμενη και συνεχης στο $ R$ συναρτηση $ f$ με συνολο τιμων το $ R$  ειναι αυξουσα στο $ Q$ θα ειναι αυξουσα και στο $ R$

Δεχόμαστε αρχικά ότι κάθε πραγματικός $ x$ είναι το όριο μιας ακολουθίας ρητών $ u_n$ (άσκηση που θα προτείνουμε στο τέλος),έτσι αν
$ x<y\Rightarrow limu_n<limv_n\Rightarrow u_n<v_n (n>n_0)\Rightarrow f(u_n)\le f(v_n)\Rightarrow limf(u_n)\le limf(v_n)\Rightarrow f(limu_n)\le f(limv_n)\Rightarrow f(x)\le f(y)$

Έστω $ x\in [a,b],a,b\in Q$ τυχαίος πραγματικός
Να δείξετε ότι $ \forall n\in N$ υπάρχει μοναδικός $ k_n\in N : 0\le k_n\le 2^n , a\plus{}k_n\frac{b\minus{}a}{2^{n}}\le x< a\plus{}(k_n\plus{}1)\frac{b\minus{}a}{2^{n}}$
Να συμπεράνετε ότι ο κάθε πραγματικός μπορεί να είναι το όριο μιας ακολουθίας της μορφής $ u_n\equal{} a\plus{}k_n\frac{b\minus{}a}{2^{n}}$ με $ 0\le k_n \le 2^n$

na rothso kati...xreiazetai pouthena oti h f exei sunolo timon to R opos anaferei i ekfonisi?

Το βιβλιο που την ειδα την εχει ετσι .Πιθανον να εννοουσε $ f: R\minus{}>R$
Η λυση μου ειναι ιδια με του Ροδολφου και δεν προσεξα αν χρειαζεται το $ f(R)\equal{}R$
Την προτεινα στα σχολικα (αν και θεωρω οτι ειναι παρατραβηγμενη για αυτο το επιπεδο) γιατι το σχολικο στην Β Λυκειου οριζει την εκθετικη με αρρητο εκθετη ως ακολουθια ρητων .. κ.λ.π.
Η αφορμη για την ασκηση δοθηκε απο κουβεντα που ειχα με μαθητη που φλερταρει το Μαθηματικο για το Μαθηματικο χασμα μεταξυ Λυκειου και Πανεπιστημιου
Μου ζητησε μια ασκηση η οποια να ειναι μετα... το Λυκειο και πριν ....το Πανεπιστημιο (αν το περιγραφω καλα)
Αυτη σκεφθηκα εκεινη την στιγμη.....

[quote="r_boris"]Όχι και για να συμπληρώσω την άσκηση
Αν η συνάρτηση είναι [b]γνησια[/b] αύξουσα στους ρητούς και συνεχής στο R θα είναι και [b]γνήσια[/b] αύξουσα στο R?[/quote]

Αυτό είναι πιο σχολικό ;)

Έστω WLOG $ x < y$, οπότε υπάρχουν ρητοί $ p,q$ με $ x < p < q < y$. Τότε, από το πρώτο ερώτημα, $ f(x)\leq f(p) < f(q)\leq f(y)$.

Cheerio,

Durandal 1707

[quote="SorrowMan"]to oti isxiei to apo panw den exei na kanei me tin piknotita tou Q sto R? Genikotera an A ena pikno iposinolo tou B kai f afksousa sto A den tha einai i f afksousa kai sto B?[/quote]

Θέλεις και συνέχεια, αλλά ναι, αυτό ακριβώς συμβαίνει (τουλάχιστον στους χώρους όπου η κλειστότητα χαρακτηρίζεται από sequential convergence).

Cheerio,

Durandal 1707


probability
function
Support
calculus
integration
probability and stats

1. Suppose X and Y have joint probability function f(x,y) = cy, x>0 , y>0, x+2y<1

Find

a) the constant c
b) are X and Y independent?
c)P(X<=0.5, Y <=0.5)
d) marginal pdf of Y
e) P(X-Y<=0.5)

2. The black-scholes model has a log normal distribution. A random variable W also has a log normal distribution if ln(W) ~ G(mean = u, std dev = s)

a)Suppose X ~ G(u,s). Find the pdf of the lognormal R.V. W = exp{x}. Also what will be the support?
b)Find the moment generating function of X
c)Use the MGF of X to find the mean and std dev. of W.

Remember that the integral of the joint density function over all possible outcomes must come to 1.  That will help you find $ c$.  To determine if $ X$ and $ Y$ are independent, you need to check whether or not $ f_{X,Y}(x,y)\equal{}f_{X}(x,y)\cdot f_{Y}(x,y)$.


[quote="pvthuan"]It should be quite easy. I think the restriction is redundant.[/quote]

Could you please post your solution ?Thanks

Thank you, Silouan, for you nice gift. I am not busy now, so wait me some minute to think...

OK. everything seems to come in its right way,

Here is my solution (it is mechanically based on SOS method)

We have to prove that
\[\sum x^{4}+2\sum x^{3}(y+z)+2xyz\sum x \ge 7\sum x^{2}y^{2}\]
\[\Leftrightarrow \frac{1}{2}\sum (x^{2}-y^{2})^{2}+2\succ xy(x-y)^{2}\ge \succ x^{2}(y-z)^{2}\]
\[\Leftrightarrow \sum (x-y)^{2}\left( (x+y)^{2}+4xy-2z^{2}\right) \ge 0.\]
We recall $S_{x},S_{y},S_{z}$ as usual and assume that $x\ge y\ge z$. Clearly, $S_{z}\ge 0$ and
\[S_{y}+S_{z}=(x+z)^{2}+4xz+(y+z)^{2}+4yz-2(x^{2}+y^{2})\]
\[=6z(x+y)+2z^{2}-x^{2}-y^{2}\ge 6z^{2}(x+y)-(x+y)^{2}\]
because $z\ge 1$ and $x+y\le 6$ (because $x^{2}+y^{2}+z^{2}=9$).

So we have done.

[quote="pvthuan"]It should be quite easy. I think the restriction is redundant.[/quote]

The restriction is necessary, thaun, if not, the inequality is wrong.

OK, I see. At least, silouan should put
\[(x+y+z)^{2}\geq x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2}.\]
With that restriction, the presence of $x^{2}+y^{2}+z^{2}$ on the left hand side member is artificial.

[quote="silouan"]hungkhtn your solution is similar to mine . Nice work  and thank you for trying the inequality  :)[/quote]

No matter! I am glad to solve your gift. And I already present you my problem in the following link

http://www.mathlinks.ro/Forum/viewtopic.php?t=144453

You can try it.


geometry
3D geometry
prism
algebra open
algebra

I have some questions I don't understand . I hope right here have some people help me . Could you show me how to do it ? Thanks !
  Question 1: If the half-life of a certain substance is ten days, how long will it take for the substance to be reduced to 1/4096 of its origional mass ?
  Question 2: What is twice the length of the space diagonal of a rectangular prism with edge lengths of 1,2, and 2 ? ( this question I don't understand what it mean )

$Q_0$ = initial quantity
$Q_f$ = final quantity
$t = \frac{days}{10}$
$4096 = 2^{12}$

$Q_f = Q_0 (0,5)^t$
$\frac{Q_0}{2^{12}} = Q_0(0,5)^t$
$1 = 2^{12}\left(\frac{1}{2}\right)^{t}$
$1 = 2^{12}(2^{-1})^{t}$
$1 = 2^{12}2^{-t}$
$1 = 2^{12 - t}$
$12 - t = 0$
$t = 12$
$12 = \frac{days}{10}$
Thus, it will take 120 days

The other exercise, just apply a few Pythagoras and you will get this:
[img]http://img.photobucket.com/albums/v280/Furuyah/math/prisma.png[/img]

The length in question is the blue one. The twice of it is $2*3 = 6$

I hope I could help

I understand what did you write. Thank you!


Pe bazele unui trunchi de piramida oarecare se iau, punctul $\ A$ pe baza de arie $\ a$ si $\ B$ pe baza de arie $\ b$.Pe segmentul $\ AB$ se ia un punct $\ M$ astfel incat $\alpha MB=\beta MA$ si se duce prin $\ M$ un plan paralel cu bazele, care determina in trunchi o sectiune de arie $\ S$.
$\ a)$Aratati ca $\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)\sqrt S=\frac{\sqrt a}{\alpha}+\frac{\sqrt b}{\beta}$
$\ b)$Fie $\ M,N$ doua puncte situate intre $\ A$ si $\ B$ astfel incat $\ AM\cdot\ AN=BM\cdot\ BN$.Prin punctele $\ M,N$ se duc planele paralele cu bazele care determina in trunchi doua sectiuni de arii $\ S_{M},S_{N}$.Sa se arate ca $\sqrt{S_{M}}+\sqrt{S_{N}}=\sqrt a+\sqrt b$

[hide="solutie"]bazele si figurile obtinute prin sectionarea triunchiului printr`un plan paralel cu bazele sunt toate omotetice, deci raportu $S/x^{2}$ este constant, unde $S$ este aria iar $x$ este un segment fixat (adica sa fie ales la fel pentru toate).
pt a), aria sectiunilor alea depinde doar de $\alpha$ si $\beta$. fie $O$ "varful" trunchiului (obtinut prin prelungire). pot deci presupune ca $O\in AB$. tinand cont de afirmatia de mai sus, relatia ce o av de demonstrat este
$OM\bigg(\frac{1}{\alpha}+\frac{1}{\beta}\bigg)=\frac{OA}{\alpha}+\frac{OB}{\beta}$. (*)
fie $A$ intre $O$ si $B$.
atunci (*) devine :
$(OA+AM)(\alpha+\beta)=\beta OA+\alpha OB$, adevarat. (tinand cont de ipoteza)
pt b), din nou aria sectiunilor nu depinde de pozitia exacta a punctelor $A$ si $B$ ci doar de pozitia planelor acelea. consider deci $O\in AB$, si ordinea punctelor $O,\,A,\,M,\,N,\,B$.
relatia devine
$OM+ON=OA+OB$. (*)
din $AM\cdot AN=BM\cdot BN$ rezulta $AM(AM+MN)=BN(BN+NM)$, sau $(AM-BN)AB=0$. deci $AM=BN$. (**)
cum (*) este echivalent cu $AM+AN=AB$, din (**) rezulta ca este adevarata.[/hide]


Find all $x$ such that $x^{3}\equiv{53}\bmod{120}$

By CRT, we want that \[\left\{\begin{array}{l}x^{3}\equiv 5\mod{8}\\ x^{3}\equiv 2\mod{3}\\ x^{3}\equiv 3\mod{5}\end{array}\right.\]Now the residues modulo 8 of $n^{3}$ are $0,1,0,3,0,5,0,7$ (when $n\equiv 0,1,\cdots 7\mod{8}$). Therefore, $x\equiv 5\mod{8}$. Similary, $x\equiv 2\mod{3}$ and $x\equiv 2\mod{5}$. The solution to \[\left\{\begin{array}{l}x\equiv 5\mod{8}\\ x\equiv 2\mod{3}\\ x\equiv 2\mod{5}\end{array}\right.\]is $x\equiv 77\mod{120}$.

[quote="maokid7"]
edit: i think i found a mistake. $120\neq{8*5*3}$[/quote]
how does that work?

o sorry. ill fix it.  :oops:  guess i was really tired

[quote="maokid7"][quote="Kurt Gödel"]...
\[\left\{\begin{array}{l}x\equiv 5\mod{8}\\ x\equiv 2\mod{3}\\ x\equiv 2\mod{5}\end{array}\right. \]
...
[/quote]
assuming that you were correct up to this point. i solved the system but got $x\equiv{59}\mod{120}$[/quote]

Except $x \equiv 59 \pmod{120}\Rightarrow x \equiv 3 \pmod{8}$. I agree with $x \equiv 77 \pmod{120}$.

[quote="maokid7"]how would you have solved the part
$x^{3}\equiv{53}\mod{R}$
if $R$ was a "large" prime such as 73?[/quote]

There's a nice solution if $gcd(3, \phi(R)) = 1$.  In that case, let $d$ be the multiplicative inverse of $3 \bmod \phi(R)$, in other words,

$3d \equiv 1 \bmod \phi(R)$

Then

$x^{3d}\equiv x \equiv 53^{d}\bmod R$

(By Euler's Totient Theorem), which is not so hard to calculate using fast exponentiation.

Edit:  By the way, you can calculate the multiplicative inverse of $3 \bmod \phi(R)$ by solving the linear diophantine

$3d+\phi(R) e = 1$

With the Extended Euclid Algorithm.


Compute the number of sets of three distinct elements that can be chosen from the set [2 1 ,2 2 , 2 3 ,..., 2 2004 , 2 2005 ] such that the three elements form an increasing geometric progression.

[hide="My solution"]
$1004004$

If $x^a$, $x^b$ and $x^c$ form a geometric progression, then $a$, $b$, and $c$ form an arithmetic progression. I don't think this needs explaining. So, we have to look for values a, b, and c where a, b, and c form an arithmetic progression and all three values are located between 1 and 2005.

$1,2,3$ --> $2003,2004,2005$: $2003$ different progressions
$1,3,5$ --> $2001,2003,2005$: $2001$

Going down we get $1,1003,2005$: 1 progression

Our solution is thus the sum of odd integers from 1 to 2003.

Using the formula $\frac {n(a(1)+a(n))}{2}$ we get $\frac {1002*2004}{2} = 1002*1002 = 1004004$

[/hide]

Looks like a harder nats sprint round or easier nats target round.

[quote="nat mc"]That's not what I got, I'll have to compare our solutions.  I'll post my solution once more people have posted[/quote]

Yeah, I found my mistake and changed my solution accordingly.

well basically you simplify this to the set (1,2,....2005) and find the number of increasing arithmetic sequences

This reminds me of #5 here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=31516&highlight= although it's slightly different.