algebra
polynomial
quadratics
function
complex numbers

Let f(x) be a quadratic such that f(x) = x has no real roots. Prove that f(f(x)) = x also has no real roots.

Ah.. I like this problem. Though f(x) doesn't have to be a quadratic, I think you mean polynomial?

Hmm, not sure if this works.

f(x) is a quadratic with two non-real roots. It can be factored:

f(x)=(x-a)(x-b) where a and b are the two roots of the quadratic equation.

f(f(x))=((x-a)(x-b)-a)((x-a)(x-b)-b)
   = (x^2-ax-bx+ab-a)(x^2-ax-bx+ab-b)
a and b were complex numbers where Im(a)!=0 and Im(b)!=0, so while Im(-a-b) might be 0, Im(ab-a) and Im(ab-b) must always be nonzero. So we have
(x^2-(some constant)x+(some non-real constant))(x^2-(some constant)x+(some other non-real constant))

Since the two constants inside the parentheses are not real, f(f(x)) can have no real roots.

Is it necessary that f(x) have two non-real roots?

[quote="TripleM"]I think you missed the = x in each case, confuted.
But in any case, as I said, I think fierytycoon misstated it, as you could tell from both the subject being polynomials and from the fact that it works for all polynomials, not just quadratics.[/quote]
Yes, I certainly did miss the =x part. Was that there originally?

[quote="Fierytycoon"]Yeah, it would work for all polynomials. And to generalize it further, we could prove that f(f(f(....f(x)))) = x has no real roots, I think (although I can't be sure, as I didn't completely understand the solution myself).[/quote]

Yep, that is correct.

[hide]Note: I will use g(x) to represent the inverse of f(x). Note also that it is not necessarily a function, although if we rewrote it in terms of y (which we're not going to) it would be a function. Technically, it's just f(x) reflected across y=x.

Ok, so we have f(x) and x and we're given that they do not intersect. Thus, when f(x) is flipped across y=x to give us g(x), they obviously do not intersect. Thus, f(x)=g(x) has no real solutions. We now take f of both sides of the equation to get f(f(x))=x has no real solutions.[/hide]

[hide]Oh, and btw, we can't have any funky discontinuous stuff since f(x) is a nice tame polynomial. Hmm...about the generalization, if my earlier method was correct, I could now show that f(f(f(f(...(x)....)=x has no real solutions if the number of "f"s is a power of 2, because I can define a(x)=f(f(x)) and b(x)=a(a(x)) etc. and use the same method. For other things...um...I'm still thinking )[/hide]

Oh well, seeing that a couple of people now have a solution I may as well post this..

[hide]

Since f(x) = x has no real roots, f(x)>x for all x or f(x)<x for all x. In the first case, f(f(x)) > f(x) > x for all x, in the second f(f(x))<f(x)<x for all x - and you can see it easily generalises.

[/hide]

you can say that because polynomials are continuous so if it were any other way intermediate value theorem would guarantee a real root.

[quote="Fierytycoon"]Yeah, the reasoning looks good.

Engel gave a weird solution though:

If f(x) has no real roots, then either f(x)>x or f(x)<x. So either
f(f(x))>f(x)>x or f(f(x))<f(x)<x.

But why can we say that?[/quote]

Doh, you posted this a minute before me.

As Nukular said - polynomials are continuous. Just think about the graph of it - if it doesn't cross the x axis, either its always above it or always below it. As easy as that.

yea i may have said this in my soln., but this works for all continuous functions apparently. nothing special regarding polynomials...

No, I understand that part (at least if I understand what you are refering to), but what I don't understand is why given f(x)>x, we can take the function of both sides and say f(f(x)) > f(x). At least I think that's what Engel's doing...

if f(y) > y,

then let y = f(x)

so f(f(x)) > f(x)
as desired

Eh, that was simple. For some reason it seemed quite suspicious to me at first. Anyways, thanks for the clarification.


This new study and associated New York Times article has been getting a lot of attention around the world.  I wanted to know what this community thought about it.

http://www.nytimes.com/2008/10/10/education/10math.html?_r=1&partner=permalink&exprod=permalink&oref=slogin

Since Dr. Andreescu was one of the authors and because of its interest to Math Circles we have been collecting links to various articles at:

http://www.metroplexmathcircle.org

My main bone to pick with this study is that most of its data came from math contest performance. I don't think that's the most accurate method of gauging mathematical talent, because "method memorization" leading to strong performances on them seems to be on the rise. I would guess if the study were done looking at the professional world (at engineers, math research, applied positions), the conclusions would not be very different. But that's a guess; we don't have that conclusion in front of us.


Do there exist irrational numbers $a$ and $b$ such that $a>1
,b>1$ and $[a^m] $ differs $[b^n]$ for any two positive integers $m$ and $n$ ?(these brackets mean floor)

I've solved this one. Take $x,y$ of the form $ a+\sqrt{a^2-1}, b+\sqrt{b^2-1}$ for $a,b$ appropriate natural numbers. It is easy then to compute the integer parts in the problem.


[quote="cuopbientcd"]Let $f(x)$ is periodic function on $R$.There exist k#-1,+1,0 such that f(kx)=kf(x). Prove that f(x) is not continuous.[/quote]
f(x)=0 forall x in R !  :D

[quote="Rust"]$k\not =1,-1,0$. Then $k\not =T$ (T period).
From $f(kx)=kf(x) \to f(\frac{1}{k}x)=\frac{1}{k}f(x)$ and $f(k^{2}x)=k^{2}f(x)$. Therefore we can find k>1, suth that $f(kx)=kf(x)$.
It proved, that f(x) not bounded in [0,T], therefore it not continiosly (because [0,T] - compact).[/quote]

Are you sure? :D


geometry
circumcircle
search
Euler
projective geometry
cyclic quadrilateral
power of a point

Let $ABCD$ be a convex quadrilateral such that $\measuredangle DAB=\measuredangle ABC=\measuredangle BCD$. Let $O$ be the circumcenter of the triangle $ABC$, and let $H$ be the orthocenter of the triangle $ABC$. Prove that the points $O$, $H$, $D$ are collinear.

If you denote by $A',C'$ respectively the points s.t. $A'B=A'C,\ C'A=C'B,\ A'\in(BA,\ C'\in(BC$, then $D=A'C\cap C'A$. The quadrilateral $A'ACC'$ is cyclic, $O$ is the orthocenter of $A'BC'$, and $H$ is the orthocenter of $ABC$. We must show that the line connecting these orthocenters passes through the intersection of the diagonals of the cyclic quadrilateral $A'ACC'$. This is a consequence of the following more general problem (it was even shortlisted a few years ago, I think):

[color=blue]Let $ABCD$ be a cyclic quadrilateral, and let $P=AC\cap BD,\ R=AB\cap CD$. Also, let $H_{1},H_{2}$ be the orthocenters of $RAD,RBC$ respectively. Then $H_{1}H_{2}$ passes through $P$.[/color]

To prove this statement: $H_{1}H_{2}$ is the Aubert line of $ABCD$ (the line passing through the orthocenters of $RAD,RBC,SAB,SCD$, where, in addition to the points defined in my previous post, we also define $S=AD\cap BC$). It's easy to show that for all quadrilaterals $ABCD$ (not just cyclic ones) this line is the radical axis of the circles constructed on diameters $AC,BD$. $P$ is clearly on this radical axis (because $PA\cdot PC=PA\cdot PD$), so $P$ belongs to this line, Q.E.D.

[[b]Moderator edit:[/b] This is again Grobber's solution to this problem, moved from a different thread:]

Let $B_{1},C_{1}$ be the intersections of the perpendicular bisectors of $AB,AC$ with $AC,AB$ respectively. Then $D=BB_{1}\cap CC_{1},\ O$ is the orthocenter of $AB_{1}C_{1}$, and $BCB_{1}C_{1}$ is cyclic. The problem is now a particular case of G8 from ISL 1995 (discussed on the forum before, I'm sure; try a search):

A circle through the vertices $B,C$ of a triangle $ABC$ cut $AB,AC$ again in $C_{1},B_{1}$ respectively. Let $P=BB_{1}\cap CC_{1}$, and let $H,H_{1}$ be the orthocenters of $ABC,AB_{1}C_{1}$ respectively. Then $P,H,H_{1}$ are collinear.

Nice solution, Grobber. A remark:

[quote="grobber"][color=blue]Let $ABCD$ be a cyclic quadrilateral, and let $P=AC\cap BD,\ R=AB\cap CD$. Also, let $H_{1},H_{2}$ be the orthocenters of $RAD,RBC$ respectively. Then $H_{1}H_{2}$ passes through $P$.[/color][/quote]

This problem was discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=29893 .

Here is another solution of the original problem, written by me for the op02 project and slightly modified:

[color=blue][b]Problem Yugoslavia 9.[/b]
Let ABCD be a convex quadrilateral such that < DAB = < ABC = < BCD. Let O and H be the circumcenter and the orthocenter of triangle ABC, respectively. Prove that the points O, H and D are collinear.[/color]

[i]Solution.[/i] What we have to show is that the point D lies on the line OH. Of course, since the points O and H are the circumcenter and the orthocenter of triangle ABC, respectively, the line OH is the Euler line of this triangle. So we must prove that the point D lies on the Euler line of triangle ABC.

Let E and F be the midpoints of the segments AB and BC.

Let X be the point of intersection of the lines DA and BC, and let Y be the point of intersection of the lines AB and CD. The equation < DAB = < ABC can be rewritten as < XAB = < ABX; thus, the triangle AXB is isosceles, with AX = BX, and it follows that the point X lies on the perpendicular bisector of the segment AB. But the point O, being the circumcenter of triangle ABC, also lies on the perpendicular bisector of the segment AB, and further, the point E, being the midpoint of the segment AB, trivially lies on the perpendicular bisector of the segment AB, too. Hence, the three points X, O and E all lie on one line - namely, on the perpendicular bisector of the segment AB. In other words, the point O lies on the line EX. Similarly, the point O lies on the line FY. Thus, $O=EX\cap FY$.

Since the points E and F be the midpoints of the sides AB and BC of triangle ABC, the lines AF and CE are two medians of triangle ABC and pass through the centroid G of triangle ABC. Consequently, $G=AF\cap CE$.

Finally, it is trivial that $D=YC\cap XA$.

Now, applying the Pappos-Pascal theorem to the collinear points A, E, Y and the collinear points C, F, X, we obtain that the points $EX\cap FY$, $YC\cap XA$ and $AF\cap CE$ are collinear. In other words, the points O, D and G are collinear. Thus, the point D lies on the line OG. But since the point O is the circumcenter and the point G is the centroid of triangle ABC, the line OG is the Euler line of triangle ABC. Thus, we have shown that the point D lies on the Euler line of triangle ABC. And we are done.

  darij

Let the perpendicular bisectors of AB, AC meet AC, AB at U, V, respectively. Let BU, CV meet at D. The triangles $\triangle ABU,\ \triangle ACV$ are isosceles with $\angle ABU = \angle UAB,\ \angle ACV = \angle VCA,$ hence $\angle ABD = \angle CAB = \angle DCA$ and we have quadrilateral ABCD with $\angle A = \angle B = \angle C.$ Quadrilateral BCUV with the diagonal intersection D is obviously cyclic, H is the orthocenter of $\triangle ABC$ and O is the orthocenter of $\triangle AUV.$ According to the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=114683]famous problem[/url], O, H, D lie on the radical axis of circles (P), (Q) with diameters BU, CV, hence these 3 points are collinear.

[quote="darij grinberg"]

Now, applying the Pappos-Pascal theorem to the collinear points A, E, Y and the collinear points C, F, X,
  [/quote]

But the six points are not on a circle. Why can't we apply Pascal Theorem? Please help.

[quote="Davron"]please can you say the source of this problem ??

davron latipov[/quote]

It says on top there... Yugoslavia 2002

Can someone answer my question?

[quote="nguyentrang"]But the six points are not on a circle. Why can't we apply Pascal Theorem?[/quote]

I have used the [b]Pappos-Pascal theorem[/b], not the [b]Pascal theorem[/b]. It states:

[color=blue][b]Pappos-Pascal theorem.[/b] Let X, Y, Z be three collinear points, and let X', Y', Z' be three collinear points. Then, the points $YZ^{\prime}\cap ZY^{\prime}$, $ZX^{\prime}\cap XZ^{\prime}$, $XY^{\prime}\cap YX^{\prime}$ are collinear. Here, the abbreviation $g\cap h$ denotes the point of intersection of two lines g and h.[/color]

This fact is also known as the Pappos theorem (or Pappus, depending on the spelling) and proven at http://www.cut-the-knot.org/pythagoras/Pappus.shtml .

By the way, my current avatar is a distorted graphic to this theorem...

  Darij

Just a "clin d'oeil"(:wink:): in France, the graphic of [b]Pappos Theorem[/b] is the first
one to present a conjecture for the pupils beginners in Geometry, in grade 6th !(chez Belin, Paris).
 It's almost the avatar of Darij ! :roll:

why don't we use complex number for this nice problem, so the problem is straightforward and we don't need any theorem more  :wink:
 Let O be the origin so $ h \equal{} a \plus{} b \plus{} c$.  we have: $ \measuredangle ABC \equal{} \measuredangle DCB \equal{} \measuredangle DAB$ (which is in the same positive/negative) so we have the system equation:
 1) $ \frac {c \minus{} b}{\widehat{c \minus{} b}} \equal{} \frac {a \minus{} b}{\widehat{a \minus{} b}}.t$ so $ c \equal{} a*t$ so $ t \equal{} c/a$
  2) $ \frac {c \minus{} b}{\widehat {c \minus{} b}} \equal{} \frac {c \minus{} d}{\widehat{c \minus{} d}}t$
   3) $ \frac {a \minus{} b}{\widehat{a \minus{} b}} \equal{} \frac {a \minus{} d}{\widehat {a \minus{} d}}.t$
from 2 and 3 we have $ d \equal{} \frac {(a \plus{} b \plus{} c)ac}{a^2 \plus{} c^2 \plus{} ac}$
so this equality is right: $ \frac {h \minus{} o}{\widehat (h \minus{} o)} \equal{} \frac {d \minus{} o}{\widehat (d \minus{} o)}$
 so we finish the proof  :roll:


I have downloaded solutions.tex, usamts.tex, and fancyhdr.sty

I am using MiKTeK 2.4.1545 and editing my files with WinEdt v. 5.4

When I attempt to compile using PDFLaTeX, it complies without errors and I get a dialog box titled "Information" that reads:

"PDFLaTeX failed to create a pdf file.
For possible explanations start the command from the Command Prompt or read the Log file (captured by WinEdt during execution)..."

I have attached the Log file to this post, but had to change the extension to .tex to upload it: solutionslog.tex

Does anyone know what the problem is, or how I can fix it?

it would be a lot better if I'd see the file you are trying to compile, but most probably the error is caused by not placing the style file in the proper place.

The file I am trying to compile is the unedited solutions.tex file.  As per the instructions, I have placed all three files (solutions.tex, usamts.tex, fancyhdr.sty) in the same directory, then tried to compile solutions.tex

None of the files have been edited, but they are attached for convenience, with the exception of fancyhdr.sty because this file cannot be uploaded.

Okay, great!  Haha, of course that makes sense.  It's working now - Thanks!

You're welcome, and good luck in the competition.

[quote="Valentin Vornicu"]in this case the solution to your problem is simple: you have to first input all the data (username, blah) and at least one character in each of the environments \begin{solution} \end{solution}.

Otherwise they will not compile.[/quote]

Just to clarify this a bit: if you leave it blank between a \begin{solution} and an \end{solution}, then no page is printed for that solution (because why would we want a blank page there?).  So if you leave all of the solutions blank, then no pages of output are generated.

So you don't need text in [u]each[/u] of the solution environments, you just need text in at least one of them.  In practical terms, this means that if you can't make any progress on solving one of the problems, then it's OK to just leave that solution blank; don't feel that you are forced to write something for every solution.

Of course, if you do make some progress towards the solution, even if you don't completely solve the problem, then go ahead and submit it, as you can receive partial credit!


I'm having a few problems with these fairly complicated log problems all help would be appreciated.

Problem 1:

The intensity of a beam of light is reduced to a fraction of its original strength after passing through [i]x[/i] metres of mist. The fraction is given by the formula [i]m = 3 ^ (-0.07x)[/i]. Calculate the distance [i]x[/i], correct to 2 decimal places, if the intensity of the light beam is reduced by half.

Problem 2:

[i]f(x) = (ax^2) + qx and g(x) = log(base n) x  given[/i]

(a) If P(2;-1) lies on the graph g, calculate the value of n.

   I got n to be 1/2

(b) A, the turning point of the parabola, also lies on the graph g. Work out the co-ordinates of A.

   I'm completely lost there.

(c) Calculate the value of [i]a[/i] and [i]q[/i][/hide]

So in the first one, $ m$ represents the strength of the beam relative to its initial strength?

Then we desire $ m(x)\equal{}\frac{1}{2}\equal{}3^{\minus{}0.07x}$.  Seeing that our base is $ 3$, our exponent is $ \minus{}0.07x$, and our answer is $ \frac{1}{2}$, we write this as $ \log_{3}\frac{1}{2}\equal{}\minus{}0.07x\equal{}\minus{}\frac{7x}{100}\Rightarrow x\equal{}\minus{}\frac{100}{7}\log_{3}\frac{1}{2}$.  If you want an actual number, you'll have to stick that into a calculator.

[quote="MarcoMP"][quote="JRav"] and our answer is $ \frac {1}{2}$[/quote]

Could you please explain how why you say it is a half? The intensity is reduced by half, wouldn't that give 1/2[i]m[/i]?[/quote]

A beam of light decreases by $ 3^{\minus{}.07x}$, so since it needs to decrease by $ \frac{1}{2}$,

$ 3^{\minus{}.07x}\equal{}\frac{1}{2}$.

What he said is correct. To 2 decimal places, -.30

[quote="MarcoMP"]Could you please explain how why you say it is a half? The intensity is reduced by half, wouldn't that give 1/2m?[/quote]
First of all, the way you have it written, $ m$ is in the denominator so that is completely wrong.  Second, I interpret the problem to mean that $ m$ gives the strength of the beam [i]relative[/i] to its initial strength, not it's actual strength.


Obtain a formula for the local truncation error for the Euler method in terms of $ t$ and the solution $ \phi(t)$:
$ y'\equal{}t^{2}\plus{}y^{2}$

According to the answer key, the answer is $ e_{n\plus{}1}\equal{}[\bar{t}_{n}\plus{}\bar{t}_{n}^{2}\phi(\bar{t})\plus{}\phi^{3}(\bar{t}_{n})]h^{2}$ but I can't figure out how to get this.

For any local truncation error estimate to make sense, you have to assume $ h<\frac1{|y|}$. The solutions do go to $ \infty$ in finite time, and single-step explicit methods will never capture that.

The local truncation error is $ y(t\plus{}h)\minus{}y*(t\plus{}h)\equal{}y(t\plus{}h)\minus{}y(t)\minus{}hy'(t)$, where $ y*$ is the Euler estimate after one step. By Taylor's theorem, this is $ y''(t_1)\frac{h^2}{2}$ for some $ t_1\in [t,t\plus{}h]$. We calculate $ y''$ from the diferential equation:
$ y''(t)\equal{}2t\plus{}2yy'(t)\equal{}2t\plus{}2t^2y(t)\plus{}2(y(t))^3$. That's your formula. All it took was a bit of implicit differentiation.


I would like to know the solution of following problem.

Prove or disprove the following statement:
For each positive integer n there exists an integer x such that:
1. x has exactly n digits
2. every digit of x is non-zero
3. x is divisible by the sum of all its digits

I don't know how to proceed - can anybody help me?

Dapet, use the fact that 10 m -1 is always divisible 9 for any natural m; this should help you proceed. Try working with that and if you are still stuck, look for me. :)

I don't understand, give me an example please. For example for n = 4 how to find any x that satisfies all conditions described above using the fact that 10m - 1 is always divisible by 9?

Thanks.

I'm not sure I understand 3X.lich's idea either :? .

Well it shouldn't take you too long to work out the details from here: if the last n digits of a number x is divisible by 2n then 2n | x.  It's easy to see that you can construct an n-digit number with all non-zero digits such that it is divisible by 2n.  Then add some extraneous digits in front so that S(x) = 2n for some appropriately selected n.

BTW this is Myth's solution (he posed the question before); I'm not going to claim the credit for it.


topology
advanced fields
advanced fields unsolved

give a detailed proof that the circle S={(x,y)in R^2: x^2+y^2=1} is a deformation retract
of the annulus A={(x,y)in R^2 : 1/4<= x^2+y^2<=4}

thx for ur help

Just give an explicit homotopy (along radial paths).

thx for your reply.

i asked a professor teaching algebraic topology about this problem, he could not find an explicit homotopy so he suggested to solve it by using an algebraic expression.

Can we do this?
appreciate your help

What does using an algebraic expression mean in this context¿

(the homotopy is given by $(x,y,t) \mapsto \left( (1-t) \cdot x+ t \cdot \frac{x}{\sqrt{x^2+y^2}} \ , \ (1-t) \cdot y+ t \cdot \frac{y}{\sqrt{x^2+y^2}} \right)$ )


superior algebra
superior algebra unsolved

Show that Aut(Z2*Z3) is isomorphic to Aut(Z2)*Aut(Z3).

find Aut(K) and Aut(Z2*Z2) where K is the Klein four-group.

[quote]Show that Aut(Z2*Z3) is isomorphic to Aut(Z2)*Aut(Z3). [/quote]
General:
Let $ G,H$ be finite groups such that $ gcd(|G|, |H|) \equal{} 1$. Prove that $ Aut(G  \times H) \simeq Aut(G) \times Aut(H)$.


find the least positive integer n such
$\frac{1}{\sin45\sin46}+\frac{1}{\sin47\sin48}+......+\frac{1}{\sin133\sin134}=\frac{1}{\sin n}$

Well-known AIME problem that's been posted many times (you'll find it in the Contests section), and moreover I don't think it's HSB.

ya the contest section doesnt have the solution posted though ya this was an AIME 2000 II 15, i think it belongs in the moderate board..

The thread is already there, though.

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=385901#p385901

I'll just lock this since t0rajir0u has posted a link to the thread.

While it is neat, it is beyond HSB.


calculus
geometry
geometric transformation
dilation
perimeter

If a square has side length $ a$, then what is the largest equilateral triangle that can fit inside the square? Prove your answer.

Stop posting one problem in multiple forums!!! Lock this topic.

[url]http://www.mathlinks.ro/viewtopic.php?p=1618356#1618356[/url]

PS: this problem was originally posted in the subforum anyways, in Middle School. Peterhi needs to be informed where to post appropriately.

First, mind your tone in these forums.

Second, as you obviously know, in the forum where this was originally posted, I was asked to post it here instead.  If there's a way to remove it from the other forum, please politely tell me what it is, and I'll be happy to do so.

[color=green]In the future, you need only wait for a mod to move the original topic posted.[/color]

First, assume that the first vertex of the triangle coincides with a vertex of the square.  (If this is not the case, then at most 3 sides of the square have points in common with the triangle, the vertices.  So we take the unused side and translate the triangle towards it until the closer vertex hits a corner of the square.)

Now assume that another vertex of the triangle is on a side of the square.  (If not, dilate the triangle from the common vertex until the assumption occurs; this only increases the area.)

Finally, assume that the third vertex of the triangle is on another side of the square.  (If not, slide the 2nd vertex along the perimeter of the square until the assumption occurs.  This increases the side length of the triangle and hence increases the area.)

But there is only one triangle that works with these assumptions, so it must be the triangle sought.  It is the triangle in the Geogebra applet below.  [geogebra]7dcea2f8d8a2edbef17a5c7eb800b30db6fa88cb[/geogebra]

We find the side length of the triangle, then find the area.

Let the 2nd vertex break the side up into $ x$ and $ a\minus{}x$, and let the side length of the triangle be $ s$.  We have $ x^2\plus{}a^2\equal{}s^2$ and also $ (a\minus{}x)^2\plus{}(a\minus{}x)^2\equal{}s^2$ so $ 2(a\minus{}x)^2\equal{}2a^2\minus{}4ax\plus{}2x^2\equal{}x^2\plus{}a^2$ or $ a^2\minus{}4ax\plus{}x^2\equal{}0$.  We want to solve for $ x$ in terms of $ a$, so:

$ x^2\minus{}4ax\plus{}4a^2\equal{}3a^2$

$ (x\minus{}2a)^2\equal{}3a^2$

$ x\equal{}(2\minus{}\sqrt{3})a$ because $ x$ has to be less than $ a$.

Now $ s^2\equal{}x^2\plus{}a^2\equal{}(2\minus{}\sqrt{3})^2a^2\plus{}a^2\equal{}(8\minus{}4\sqrt{3})a^2$

so the area of the equilateral triangle is $ \frac{\sqrt{3}s^2}{4}\equal{}\frac{8\sqrt{3}\minus{}12}{4}a^2\equal{}(2\sqrt{3}\minus{}3)a^2$.


Show that every real number x satisfies $ \minus{}1<\sqrt{x^2\plus{}x\plus{}1}\plus{}\sqrt{x^2\minus{}x\plus{}1}<1$

You meant $ \minus{} 1 < \sqrt {x^2 \plus{} x \plus{} 1} \minus{} \sqrt {x^2 \minus{} x \plus{} 1} < 1$ ?  :)

[hide="Solution"]If we show that $ 0 < \sqrt {x^2 \plus{} x \plus{} 1} \minus{} \sqrt {x^2 \minus{} x \plus{} 1} < 1$ for positive $ x$, then substituting $ x$ with $ \minus{} x$ proves that the negative numbers satisfy the equation as well.

[b]The 1st Part:[/b] For positive $ x$, it is enough to note that $ x^2 \pm x \plus{} 1 > 0$, and thus:
$ x^2 \minus{} x \plus{} 1 < x^2 \plus{} x \plus{} 1$
$ \implies \sqrt {x^2 \minus{} x \plus{} 1} < \sqrt {x^2 \plus{} x \plus{} 1}$
$ \implies 0 \sqrt {x^2 \plus{} x \plus{} 1} \minus{} \sqrt {x^2 \minus{} x \plus{} 1}$

[b]The 2nd Part:[/b]

$ \text{\color{blue}{Trivially: }} \qquad 4x^4 \plus{} 4x^2 \plus{} 1 < 4 x^4 \plus{} 4x^2 \plus{} 4$

$ \text{\color{blue}{Take squareroot: }} \qquad \implies 2x^2 \plus{} 1 < 2 \sqrt {x^4 \plus{} x^2 \plus{} 1}$

$ \text{\color{blue}{Rearrange: }} \qquad \implies (x^2 \plus{} x \plus{} 1) \plus{} (x^2 \minus{} x \plus{} 1) \minus{} 2 \sqrt {(x^2 \plus{} x \plus{} 1)(x^2 \minus{} x \plus{} 1)} < 1$

$ \text{\color{blue}{Take squareroot: }} \qquad \implies \sqrt {x^2 \plus{} x \plus{} 1} \minus{} \sqrt {x^2 \minus{} x \plus{} 1} < 1$[/hide]

[hide="Alternately"] Equivalently, $ |2x| < \sqrt{x^2 \plus{} x \plus{} 1} \plus{} \sqrt{x^2 \minus{} x \plus{} 1}$.  Squaring,

$ 4x^2 < x^2 \plus{} x \plus{} 1 \plus{} x^2 \minus{} x \plus{} 1 \plus{} 2 \sqrt{x^4 \plus{} x^2 \plus{} 1}$.

But $ x^4 \plus{} x^2 \plus{} 1 > x^4$, so the RHS is greater than $ 4x^2 \plus{} 2$. [/hide]

The given is equivalent to $ \sqrt {(x \plus{} 1/2)^2 \plus{} 3/4} \minus{} \sqrt {(x \minus{} 1/2)^2 \plus{} 3/4}$. But putting a 3/4 under both radicals only makes the difference smaller, as it dilutes the value of the rest.

If this is not rigorous enough for you, here is a proof that $ \sqrt {a \plus{} x} \minus{} \sqrt {b \plus{} x} \le \sqrt {a} \minus{} \sqrt {b}$
Squaring, $ a \plus{} b \plus{} 2x \minus{} 2\sqrt {(a \plus{} x)(b \plus{} x)} \le a \plus{} b \minus{} 2\sqrt {ab}$
$ x \plus{} \sqrt {ab} \le \sqrt {(a \plus{} x)(b \plus{} x)}$
$ x^2 \plus{} ab \plus{} 2x\sqrt {ab} \le x^2 \plus{} ax \plus{} bx \plus{} ab$
$ \sqrt {ab} \le (a \plus{} b)/2$
Which is true by AM-GM.

This inequality can also be proven using geometry.  We assume $ x$ is positive, since if is negative, make the transformation $ x\to\minus{}x$ and you get the same inequality.

Now, as in the picture, let $ AD\equal{}x,BD\equal{}1,CD\equal{}1,\angle ADB\equal{}60,\angle BDC\equal{}60$. Thus, we have $ AB\equal{}\sqrt{x^2\minus{}x\plus{}1},AC\equal{}\sqrt{x^2\plus{}x\plus{}1},BC\equal{}1$.  From the triangle inequality we have $ AC\le BC\plus{}AB$ or

$ \sqrt{x^2\plus{}x\plus{}1}\minus{}\sqrt{x^2\minus{}x\plus{}1}\le1$

Equality occurs when $ A,B,C$ are collinear.  But this never happens since $ AD\parallel{}BC$.  Thus we have for all $ x$,

$ \minus{}1<\sqrt{x^2\plus{}x\plus{}1}\minus{}\sqrt{x^2\minus{}x\plus{}1}<1$

[hide="Remark"] Perhaps only dgreenb801's solution suggests strongly that the inequality is sharp; in other words, $ 1$ cannot be replaced by a smaller constant.  Indeed,

$ \lim_{x \to \infty} \sqrt{x^2 \plus{} x \plus{} 1} \minus{} \sqrt{x^2 \minus{} x \plus{} 1} \equal{} \lim_{x \to \infty} \frac{2x}{\sqrt{x^2 \plus{} x \plus{} 1} \plus{} \sqrt{x^2 \minus{} x \plus{} 1}} \equal{} 1$. [/hide]


integration
calculus
calculus computations

Hi I was wondering if anyone can prove this:

If $ \displaystyle s > 0$ and $ n > 1$ that
$ \displaystyle\mathcal{L}\left\{\frac {t^{n \minus{} 1}}{1 \minus{} e^{ \minus{} t}}\right \} \equal{} \Gamma (n)\left\{\frac {1}{s^n} \plus{} \frac {1}{(s \plus{} 1)^n} \plus{} \frac {1}{(s \plus{} 2)^n} \plus{} \cdots \right\}$

Expand the former as a series, and show that the Laplace transform of $ t^{n\minus{}1}e^{\minus{}kt}$ is $ \Gamma(n)(s\plus{}k)^{\minus{}n}$. That's basically just a scaling substitution to get the standard $ \Gamma$ integral.

Wow, I did not know that it was that simple.  Thank you jmerry.

$ \displaystyle\mathcal{L}\left\{\frac{t^{n\minus{}1}}{1\minus{}e^{\minus{}t}}\right\}\equal{}\mathcal{L}\left\{t^{n\minus{}1}\sum_{k\equal{}0}^\infty(e^{\minus{}t})^k\right\}\equal{}\sum_{k\equal{}0}^\infty\mathcal{L}\{t^{n\minus{}1}\cdot e^{\minus{}kt}\}$
Since $ \displaystyle \mathcal{L}\{t^{n\minus{}1}\cdot e^{\minus{}kt}\}\equal{}\int_0^{\infty}e^{\minus{}st}(t^{n\minus{}1}e^{\minus{}kt})dt$, let $ u\equal{}(s\plus{}k)t$
Then $ \displaystyle \mathcal{L}\{t^{n\minus{}1}\cdot e^{\minus{}kt}\}\equal{}\frac{1}{(s\plus{}k)^n}\int_0^\infty u^{n\minus{}1}e^{\minus{}u}du\equal{}\frac{\Gamma (n)}{(s\plus{}k)^n}$

Hence, $ \displaystyle\mathcal{L}\left\{\frac{t^{n\minus{}1}}{1\minus{}e^{\minus{}t}}\right\}\equal{}\Gamma (n)\sum_{k\equal{}0}^\infty \frac{1}{(s\plus{}k)^n}$


[quote="topofmath"]Does anyone know any websites that post Mathcounts 1990-2001 and national?
Thanks.[/quote]

Actually, putting free Mathcounts materials online is illegal. You must trade with other people or buy the materials from mathcounts off Mathcounts.org


A small software company publishes computer games and educational and utility software. Their policy is to market a total of 36 new programs each year, with at least four of these being games. The number of utility programs published is never more than twice the number of educational programs. On average, the company makes an annual profit of 5000 dollars on each computer game, 8000 dollars on each educational program and 6000 dollars on each utility program. How many of each type of software should they publish annually for maximum profit?

Can anybody help me with this, please?

Clearly, computer games are worth the least, so the company will want to produce the minimum possible number of computer games, or $ 4$. Next, the company will prefer educational programs to utility programs, so the other $ 32$ programs should all be educational programs. (Note that the second condition is satisfied.)

The company should publish $ 4$ computer games and $ 32$ educational programs.


Let $F_k$ denote the $k$th fibonacci number and $(a,b)$ denote the greatest common divisor of $a$ and $b$. Given positive integers $m$ and $n$ and that $(F_m,F_n)=F_{(m,n)}$, prove that $F_m$ divides $F_n$ if and only if $m$ divides $n$.

If $m$ does not divide $n$, then $(m,n)=d$ where $d<m$
then $(F_m,F_n)=F_d$ and $F_d<F_m$

[quote="Andreas"]$(m, n) = m$. Hence $(F_{m}, F_{n}) = F_{m}$[/quote]

i don't understand how you make that step

You are given that $(F_m, F_n) = F_(m,n)$. After assuming $m | n$, then $(m,n) = m$, this is obvious.


The reason I want to know is that although this is a fairly simple result, I've seen a disproportionate number of problems that would be trivialized by its invocation... so anyway, here goes:

[i]Given a finite state machine in which any state uniquely determines both its preceding and succeeding states, the machine must enter a cycle beginning with the start state.[/i]

The proof is fairly simple. First, anytime we return to a state we've been to before we must enter a cycle. Since there are only a finite number of states, this must happen. Second, suppose from the start state we have the following sequence of steps, with $ S_n$ as the first state reached that is included in the eventual cycle:
$ S, S_1, S_2, ... , S_{n \minus{} 1}, S_n, S_{n \plus{} 1}, ... S_{n \plus{} m}, S_n, ...$
i.e., the machine enters a cycle that does not include the start state. Notice that since preceeding states are uniquely determined, only one state preceeds $ S_n$, so $ S_{n \plus{} m} \equal{} S_{n \minus{} 1}$, contradicting minimality of $ S_n$. Thus, any cycle must return to the start state as desired. $ \square$

I don't think that such a simple theorem needs a name, most proofs would just state this is a trivial fact :D. At least if computer science proofs are anything like mathematical proofs. Just out of curiosity, where would you use this result?

Well, it actually happened to be in a graph theory problem ;) (well sorta... the problem wasn't about any property inherent to the graph, but rather to it's representation...)

[quote]If you alternate between taking the leftmost and the rightmost route at every junction in a maze, are you guaranteed to get back to the place where you started out?[/quote]

There was also a note stating that a maze was considered to be a finite graph with $ deg(v) \geq 3$ for all $ v \in V$

I guess you're right about it being too simple to have to prove, but it would be a lot easier to just say "by the XYZ theorem" than "since we can view the process as a finite state machine with blah blah blah..."


Given a right triangle $\triangle PXY$ with right angle at $X$, form a new right triangle $\triangle PX'Y'$ that has acute angle $\angle P$ in common with the given triangle but double the hypotenuse.
Prove that the side opposite the acute angle is at least doubled, whereas the side adjacent to the acute angle is at most doubled.

It seems that your problem needs clarification.

Put $X'$ on $\vec{PX}$ with $PX'=2PX$, and $Y'$ on $\vec{PY}$ with $PY'=2PY$. Then $\triangle PXY \sim \triangle PX'Y'.$ All three sides exactly doubled.


Burning butane (C4H10) produces gaseous carbon dioxide and water.  The enthalpy of combustion of butane is -2650 kJ/mole.  Determine how much water you can heat from room temperature (22 deg C) to boiling with 1 lb of butane.

Thanks in advance!

$ 1\,lb \equal{} 453,59237 \,g$, and the molar mass of butane is $ 58.123\,g/mol$, and so $ 1 \,(butane) \equal{} 7.80 \,mol$. Through the combustion of this chemical quantity of butane there is the transfer of $ 7.80 \,mol \times 2650\, kJ/mol \equal{} 2.07 \times 10^7 \,J$ as heat to the surroundings. If the surroundings is water, then it will receive heat, and considering that the water's heat capacity remains constant on this process, we are allowed to use the equation $ Q \equal{} mc \Delta T$:

$ 2.07 \times 10^7 \,J \equal{} m \cdot 4.1813\, J/g/K \cdot (100 \minus{} 22)$

$ \equal{}> m \equal{} 63.5 \,kg$


trigonometry
induction
irrational number
algebra unsolved
algebra

Maybe posted before...
\[f(x+y)+f(x-y)=2f(x)\cos{y}\]

Simple indeed.
The general set of solutions are f(x)= a sinx where a is any real number. In particular f(x)=0 is a solution.

Now I shall proceed to prove this.  First let f(0)=a.
Then f(x)+f(-x)=(2cosx)a
also f(2x) +a = 2f(x)cosx
Now 2acos2x = f(2x)+f(-2x) = 2cos2x(f(x)+f(-x)) -2a = a(4cosx cos2x -2)
so a(2cos2x -4cosx cos2x +2)=0 thus we obtain that a=0. hence f(0)=0
i) f(x)=-f(-x)
ii)f(2x)=2f(x)cosx
Now f(3x) + f(x) = 2f(2x) cos x = 4f(x)cos^2 x so f(3x)=f(x) (4cos^2x - 1)
Case1. f(x)=0 for all rational numbers x. so f(x)=0 for all x. This actually requires a bit of work...
case2. f(x) is not zero for some rational. Then you can see that if f(r) is not zero.
f(2r)/sin2r =f(r)/sinr and similarly f(3r)/f(r)=f(r)/sinr ... indeed you can use induction to show that
f(nr)/f(r)=f(r)/sinr ... or with a bit more work that f(qr)/sinqr= f(r)/sinr =a for all q rational.
So we get f(q)= asinq for all rational numbers q.
I post the second part of solution later.
(1)
-A

Second part:
let x be an irrational number and r a rational.
then f(x+r) +f(x) = 2f(r/2) cos r/2
as you can see the exact approach as above yields that f(x+r) = a(x) sin (x+r)
for all rational r. [note that x is fixed]

Let x and y be two general irrational numbers.
f(x+r) + f(y+q) = 2f( (x+y)/2 + (r+q)/2) cos( x-y)/2 + (r-q)/2)
Now replacing f(x+r) with a(x) sin (x+r) ,...
we get an equation that needs to be satisfied by a(x), a(y) and a(x+y)/2. but for all rationals.
this implies that there is at most one solution. (of course this is not rigorous. One has to work a bit more on this...)
Finally a(x)=a(y)=a is clearly a solution. So the general solution is f(x)=asinx.
(2)
-Ali

[quote="shobber"]thanks.
it is from my book. The answer given by the author is $f(x)=a\sin{x}+b\cos{x}$, but i think it is wrong.[/quote]
i made some mistakes reading my book's solution, but i've made some corrections.

Question: Solve $f(x+y)+f(x-y)=2f(x)\cos{y}$
Let $x=0$ and $y=t$, so we have:
\[f(t)+f(-t)=2f(0)\cos{t}......(1)\]
Let $x=\frac{\pi}{2}+t$ and $y=\frac{\pi}{2}$, so we have:
\[f(\pi+t)+f(t)=0......(2)\]
Let $x=\frac{\pi}{2}$ and $y=\frac{\pi}{2}+t$, so we have:
\[f(\pi+t)+f(-t)=2 f(\frac{\pi}{2}) \cos{(\frac{\pi}{2}+t)}......(3)\]

$(1)+(2)-(3)$, and let $f(0)=b$ and $f(\frac{\pi}{2})=a$, so we have:
\[f(x)=a\sin{x}+b\cos{x}\]

[quote="ali"]very nice solution,(at least much shorter than mine) but one step from being complete.
b=0. indeed i proved in the beginning of my solution that f(0)=0.
-Ali[/quote]
Can you please explain this?? I don't quite understand that step...

my mistake...
your solution is perfectly fine. :D
In fact I also get the soltuion you got.(I missed cosx solution's originally)


geometry
geometric transformation
homothety
linear algebra
matrix
linear algebra unsolved

let $P$ be a complex polynomial.
$E=\{M \in M_n(C) s.t. P(M)=0\}$.
Prove that any isolated point of $E$ is an homothety

Since every conjugate of a matrix in $E$ is also in $E$, it suffices to prove this:

If $A$ is not a homothety, there are invertible matrices arbitrarily close to $I_n$ which do not commute with $A$.

We may as well assume $A$ is in Jordan form. If it's not diagonal (say it has a $1$ in position $(2,1)$), then we can find diagonal matrices with entries as close to $1$ as we wish, which will certainly not commute with $A$ as long as they have different first and second entries on the diagonal, for example. If $A$ is diagonal, on the other hand, and has, say, distinct first and second entries, then we use matrices which have $n$ equal entries on the main diagonal, all very close to $1$, an entry very close to $0$ in position $(2,1)$, and $0$ everywhere else (i.e. we sort of "reverse" what we did above for non-diagonal matrices).

Once we have shown the above, we can take a sequence $A_k\to I_n$ of invertible matrices s.t. $A_kAA_k^{-1}\ne A$. Clearly, $A_kAA_k^{-1}\to A$, and $A$ is an accumulation point of $E$.


inequalities
combinatorics proposed
combinatorics

Prove the following identity:
$\sum_{i=0}^n(-1)^iC_n^iC_{m+n-i}^{k-i}=C_m^k$ for every $m \geq k\geq 0, n\geq 0  $.
Try to find a combinatorics using solution.

We will use the classical convention  that $C_a^b = 0$ if $b>a$.

Let's consider a population of $m$ girls and $n$ boys. We want to select a group of $k$ people, and among them choose some leaders (perhaps none) but leaders have to be boys.
For a number $i$ of leaders, we may first choose the $i$ leaders from the $n$ boys then complete the group by any $k-i$ people from the remaining $m+n-i$.
Thus, the left-hand side of the desired identity is just the total number of group with an even number of leaders minus the total number of groups with an odd number of leaders.

Now, we may form a group in another way : First, we select the $k$ people from the $m+n$ ones and then we choose the leaders from the $g$ boys which belong to this group. Therefore, from a given group we may choose $\sum_{i \geq 0} C_g ^{2i}$ groups with an even number of leaders and $\sum_{i \geq 0} C_g ^{2i+1}$ groups with an odd number of leaders. But, it is very well-known that if $g>0$ these two sums are equals (there can be proved by a combinatorial interpretation or by the binomial theorem). Thus, the chosen groups of $k$ people which give a non-zero contribution in the LHS of the desired inequality are only those which do not contain any boys. It follows that the LHS is equal to the number of groups of $k$ people from the $m$ girls (and then with no leaders), that is $C_m^k$.

Pierre.


What do you think is the minimum score you need to get on USAMO for a freshman if you want to make the MOSP?

I imagine that a 1 will probably do it, but maybe shoot for a 4-ish?

I imagine that there will be several freshman $0$s.  If you solve one problem, you're a shoe-in.  Not so easy for me...  :(  I probably need about 20...

Freshman here, and I'm aiming for a two or three. Should be enough to qualify though, I hope...

Freshman also.  I think 4 would be definitely enough to qualify.  I'm aiming for a 20ish though.  At the least I hope to beat my score last year.  If I didn't, it would be really sad, unless this USAMO is ridiculously hard.

What is the best possible score? And how does the grading work?

42 -- 7 points per question, 6 questions. Grading is kind of complicated and subjective because they are proofs.

There is a rubric for each problem so it's less subjective than one might think. Generally, if you make decent progress or have a partial solution, you can expect 1 or 2 points. 5 and 6 points are generally when you've got everything except some minor details. Anything in between might be 3 or 4 points. And, of course, 7 is a complete proof.

Thanks, I hope I can make it to the USAMO my next year.

[quote="MithsApprentice"]I imagine it's not impossible that all frosh are invited. A 1 or 2 would be nice though...[/quote]

I like that. I hope that all 38 freshmen get invited. Just work hard on the questions and think positively.

[quote="winluo2"]20!?!? That's crazy... what'd you get last year?[/quote]

Hey, I didn't say it was realistic :D.  But if this USAMO is easy, and some other fortuitous factors come in, it would be possible (for a lot of us I think).


algebra
polynomial
number theory proposed
number theory

Find all polynomials $f$ with integer coefficient such that, for every prime $p$ and natural numbers $u$ and $v$ with the condition:
\[ p \mid uv - 1 \]
we always have $p \mid f(u)f(v) - 1$.

[quote="TTsphn"][quote="khashi70"]Find All Polynomials $ f$ with integer coefficient such that , for every prime $ p$ and every natural numbers $ u$ and $ v$ with the condition :
$ p|uv \minus{} 1$
we always have $ p|f(u)f(v) \minus{} 1$[/quote]
A good problem ! .If f is constant then it's easy to show that f = 1 or f = - 1 . If not ,assume that f has degree n . Let  $ g(x) \equal{} x^n.f(\frac {1}{x})$ , then g(x) is a  polynomial with integer coefficients . We will show that for all  prime $ p$ then $ p|f(x).g(x) \minus{} x^n$ . By the condition we have $ p|f(u)f(u^{ \minus{} 1}) \minus{} 1$ for all positive integer $ u$  relative prime to $ p$ integer  therefore $ p|u^n.f(u)f(u^{ \minus{} 1}) \minus{} u^n$ or $ p|f(u)g(u) \minus{} u^n$ for all $ \gcd(u,p) \equal{} 1$
Now for a fixed integer $ u$ we take p very large ($ p > |\max\{|f(u).g(u) \minus{} u^n|,u$) then by the condition we must have $ f(u).g(u) \minus{} u^n \equal{} 0$ .This equality holds for all natural numbers n so by the property of polynomial we obtain :$ f(x)g(x) \equal{} x^n$ for all real $ x$ . Because degree of $ f$ is n so $ g$ must be constant ,which means that $ f(x) \equal{} a.x^n$ and it's not very hard to show that $ a \equal{} 1$ or $ a \equal{} \minus{} 1$ .
Thus all solutions of this problem are $ p(x) \equal{} ax^n$ where $ n\geq 0$ and $ a\in\{ \minus{} 1;1\}$ .[/quote]
hi,
 $ u$ and $ v$ must be natural numbers so when you put $ v \equal{} u^{ \minus{} 1}$then $ u \equal{} 1$.

$ u^\minus{}1$ is the natural number $ v$ such that $ uv\equal{}1(\mod p)$ . It is not $ \frac{1}{u}$

yes it's clear with $ Z/pZ$, but in the first time when i look an expression like $ g(x) \equal{} x^nf(\frac {1}{x})$ Ithink that you mean $ \frac {1}{u}$in your reply .

maybe I'm confused with -1 signs and 1/x somewhere so i have one question if its not a problem for someone to answer doesn't u-1 depends on p except u so how can he be sure that with growth of p u-1 doesn't overgrow it?

thanks in forward
regards

I'm silly I got it thanks anyway, very nice solution :)

The following general problem is also true ( I believe that !)
Let $ f$ be a polynomial with degree greater than or equal . Let g be a polynomial satisfying condition: for any prime divisors of $ f(u)f(v)\minus{}1$ then $ p|g(u)g(v)\minus{}1$.Prove that $ g(x)\equal{}f(x)^k$ or $ g(x)\equal{}\minus{}f(x)^k$ for some positive integer $ k$

[quote="khashi70"]Find All Polynomials $ f$ with integer coefficient such that , for every prime $ p$ and every natural numbers $ u$ and $ v$ with the condition :
$ p|uv\minus{}1$-
we always have $ p|f(u)f(v)\minus{}1$[/quote]

Suppose $f(x)=\sum a_ix^i$ and $deg(f(x))=n$.Now $f(u)f(v)-1=\sum a_ia_ju^iv^j\equiv \sum a^2_i+g(u)-1(modp)$ where certainly $g(x)$ is a polynomial of degree less than $n$ with fixed degree and coefficients.Now it's is well know that if $p|P(x)$ for all $x\in\mathbb N$ with $deg(P)<p$ then $p$ divides all coefficients of $P$.So from that we get for all prime, $p>n$ we've $p$ divides all coefficients of $g$ and that implies all of them are zero. So now we must have for all prime $p>n$ the fixed expression $p|(\sum a^2_i)-1$ and basically that implies $a^2_n=1,a_{n-1}=a_{n-2}=.....=a_1=0$ , so finally we get $f(x)\pm x^n$ for all $x\in\mathbb Z$.

[hide="another solution"]suppose $p\mid f(x)$ and $(p,x)=1\Rightarrow \exists x^{\ast }:\: xx^{\ast }\equiv 1\: (mod\: p)$
contradiction !!

so if $p\mid f(x)\Rightarrow p\mid x$ so we have :

$q\mid f(q)$ for any prime number . so $f(x)=xg(x)$
 [/hide]

Okay, a nice one but a bit easy for Iran I guess.

Firstly, $f(x)=cx^n$ where $c$ is either of $+1,-1$ is a solution and we claim it to be the only one.
The idea is natural:
(Considering polynomials over the ring with coefficients in residue classes modulo $p$)
We note that if $f(x)=a_nx^n+...+a_1x+a_0$ where $a_n$ is non-zero.
And let $n$ be the least positive integer for which solutions other than those mentioned exist.
Then clearly, $a_0$ cant be $0$ or else we trivially reduce the degree.

Now, it comes down to the fact that
$p \mid g(x)= (a_nx^n+...+a_1x+a_0)(a_0x^n+...+a_{n-1}x+a_n)-x^n$ $\forall x \in \mathbb{Z}$.

Now, take $x=1,2,...,2n+1$ and let $p$ be a prime greater than the largest of the absolute values $g(1),...,g(2n+1)$.
Then it is evident that $g(1)=...=g(2n+1)=0$ and $g$ has degree $2n$ so $g \equiv 0 \forall x\in \mathbb{Z}$.
This indeed, gives that the leading coefficient of $g$ is $0$,i.e, $a_na_0=0$. But, both by assumption are non-zero giving the desired contradiction.

[quote=TTsphn]

therefore $ p|u^n.f(u)f(u^{ \minus{} 1}) \minus{} u^n$ or $ p|f(u)g(u) \minus{} u^n$ for all $ \gcd(u,p) \equal{} 1$
[/quote]

Sorry for digging this up but I can't understand this bit. By definition of $g(x)$, $g(x)=x^nf(\frac{1}{x})$. But here $u^{-1}\ne \frac{1}{u}$

The only constant solutions are $\pm 1$ and $f$ is a solution iff $-f$ is a solution hence assume the leading coefficient to be positive and $\mathrm{deg}(f) \ge 1$. Say we have a prime $q \neq p$ such that $q \mid f(p)$, by Fermat's Little Theorem we can pick $\alpha$ such that $q \mid p\cdot p^{\alpha} - 1 \implies q \mid f(p)f(p^\alpha) -1$, thus
$$0 \equiv f(p) \equiv f(p)f(p^{\alpha}) \equiv 1 \pmod q$$
forcing a contradiction, hence $p \mid f(p)$ and since it holds for infinitely many $p$'s it implies $f(0)=0$ and we can write $f(x)=xg(x)$ and perform the same operation on $g$ until we eventually reach a polynomial of degree $0$. This concludes that the solution set for $f$ is $\{x^n , -x^n\}$ for any fixed non negative integer $n$.


number theory
prime numbers
number theory unsolved

Let $ a_1,a_2,...$ infinite sequence of natural numbers such that every prime numbers not divide all $ a_1,...$
Prove that there exists subsequence $ a_{i_1},a_{i_2},...$ for $ 1\leq i_1<i_2<...$ such that $ \gcd(a_{i_j},a_{i_k}\equal{}1$ for all $ j\neq k$

This can't be true : let $ a_1 \equal{} 2$ and for all $ k > 1$ let $ a_k \equal{} 3b_k$ for some positive integer $ b_k$.
You won't find any infinite subsequence of coprime numbers... :wink:

I think you have to assume that each prime divides only a finite number of terms
Or that for each prime $ p$ there are infinite terms which are not multiples of $ p$.

Pierre.


Stanford college AMC USA(J)MO USAMO Princeton

For regular master's degrees, not at all.  That's partly because math people who intend to get Ph.D.s don't usually enroll in Masters programs (you have to pay tuition in Masters programs but usually not Ph.D. programs).  For professional science masters degress, I don't know.

In terms of Ph.D. admissions, the undergraduate institution plays a role pretty much only in the sense of what's offered at that institution.  For example, if you attend a school which offers one semester of algebra and one of analysis (as opposed to two) then you'll have a tougher time than if you go to a school which offers year-long sequences.  (You can mitigate this by taking independent study courses in the remaining material, but that in turn depends on whether there are faculty willing to supervise you.) It also depends on what the curriculum is in those courses; many graduate schools ask you to list the textbook title/author/edition for each math course as part of their application process.

Other than that, it's what you do with what's offered, and what other experiences you have (REUs and the like) that makes the most difference.

For PhD programs in the sciences, faculty recommendations are extremely important.  I wouldn't be surprised if they're the determining factor in many applications.  When a prof takes on a grad student, she/he is hiring a worker they'll have for 5-6 years.  They are going to take the word of a trusted friend at another institution very seriously.

(On a personal note, I applied to grad school a couple months after the deadline, and only after one of my professors found out I had no plans and gave me a list of schools I should apply to.  In other words, places he knew people at.  I picked the California schools (Stanford, Berkeley, CalTech), and got into all three despite sending in an incomplete application 2 months late.  The recommendation, I think, was crucial.)

As I remember, he wasn't applying to graduate school in mathematics...

What did he apply then? Not biochemistry, I assume. If it was economics or finance, it would only make him more appealing.

[quote="rrusczyk"][quote="qvc122"]rrusczyk, in your case, it was really your unusually outstanding qualification, which brought you acceptance to all three schools. The recommendations were probably just coincidence. Imagine, being an USAMO winner, multiple trips to MOP, graduated from Princeton, where could you possibly find one like that? Nevertheless, recommendations for graduate school (and for jobs after graduate school), especially from someone inside the circle are extremely useful.[/quote]

The grad school was in chemical engineering.  MOP and USAMO didn't mean much for that.  (I looked into applying to math grad schools, but it was made very clear to me that I wouldn't get into any worth going to.)  It was made subtly clear at the time, and more explicitly clear later that the recommendations were the key to being able to apply so late and still have a pick of schools.[/quote]

why wouldnt u have been able to get into any worth going to?
also is it possible to go straight from being an undergrad. into a Ph.D program?

[quote="maokid7"]also is it possible to go straight from being an undergrad. into a Ph.D program?[/quote]

Yes, usually mathematicians apply directly from college to Ph.D. programs.

[quote="maokid7"]how much dose where you attened school for you undergraduate degree play into where you will be able to go to attain you Masters or even PhD?[/quote]

From what I understand, most people who do Master's programs are the ones who did undergrad at the same place and forgot to apply/didn't get in anywhere they liked and stuck around for another year.  There are also some who apply for the PhD program for the free tuition + stipend and then leave after they get their Master's in the first year, but I think that's rather rare.


algebra
polynomial
analytic geometry
abstract algebra
projective geometry
advanced fields
advanced fields unsolved

Hi all,

If $ Y \subseteq \mathbb{P}^2$ is a curve defined by a homogeneous polynomial $ f$, one can regard the set of lines in $ \mathbb{P}^2$ as another projective space $ (\mathbb{P}^2)^*$, by taking $ (a_0, a_1, a_2)$ to be homogeneous coordinates of the line $ L: a_0x_0 \plus{} a_1x_1 \plus{} a_2x_2 \equal{} 0$.

I'm supposed to show that for each $ P \in Y$ nonsingular, there is a unique $ T_P(Y) \in (\mathbb{P}^2)^*$ whose intersection multiplicity with $ Y$ at $ P$ is $ > 1$.  To show uniqueness, does this work?  The notation $ T_P(Y)$ suggests that I consider the mapping $ T: Y \rightarrow (\mathbb{P}^2)^*$ sending $ P \mapsto T_P: (\partial f/\partial x_0)(P) \cdot x_0 \plus{} (\partial f/ \partial x_1)(P) \cdot x_1 \plus{} (\partial f/ \partial x_2)(P) \cdot x_2$.  This map is injective, right?  ...which would prove uniqueness?  The intersection multiplicity is defined to be the length of the $ \mathcal{O}_P$-module $ \mathcal{O}_P/(f,T_P)$.  Should this be equal to the degree of the curve $ Y$ (hence $ > 1$)?  Actually, how can I find the length of this module?

Finally, the map $ P \mapsto T_P(Y)$ defines a morphism of $ \mbox{Reg } Y$ (the set of nonsingular points of $ Y$) into $ (\mathbb{P}^2)^*$.  This part seems easier, since the map is well defined as long as one of $ (\partial f/\partial x_i)(P)$ is nonzero, i.e., $ P$ is nonsingular.

Can anyone point out potential flaws/inaccuracies/things that need further proof?

Sorry, another question:

Ex. I.5.4 in Hartshorne defined the intersection multiplicity of two affine curves $ Y,Z \subseteq \mathbb{A}^2$ at a point $ P \in Y \cap Z$ to be the length of the $ \mathcal{O}_P$-module $ \mathcal{O}_P/(f,g)$.  Can one define similarly intersection multiplicity for curves in $ \mathbb{P}^2$?  For a curve $ Y$ of degree $ d$ in $ \mathbb{P}^2$ and a line $ L \subset \mathbb{P}^2$, $ L \neq Y$, the same exercise defined intersection multiplicity to be $ (L \cdot Y) : \equal{} \sum (L \cdot Y)_P$ taken over all $ P \in L \cap Y$, where $ (L \cdot Y)_P$ is defined using a "suitable" affine cover of $ \mathbb{P}^2$.  What is meant by "suitable" in this case?  Since $ Y$ and $ T_P$ should only intersect at the point $ P$, what even constitutes an unsuitable cover?


geometry
geometric transformation
reflection
IMO
IMO 2005

G35
S23
B12

Oh, and a Moldova contestant got a special prize for a really nice solution for Q3.

Dear [b]Yossarian[/b] are these numbers official?

As far as i know, yes, these cut-offs are official.

The cut-off mentioned above reflects how the medal distributed among the contestants. See here:

http://erdos.fciencias.unam.mx/results.htm

It follows that the cut-off above is official  :D  :D  :D

[quote="Megus"][quote="Quique"]It is oficial.
I was in the jury meeting when they decided the distribution.
There will also be a special solution prize.[/quote]

Can anybody show this solution ?  :?[/quote]

\[\frac{x^5 - x^2}{x^5 + y^2 + z^2} - \frac{x^5 - x^2}{ x^3(x^2+y^2+z^2} = \frac{(x^3-1)^2 x^2(y^2+z^2)}{x^3(x^2+y^2+z^2)(x^5+y^2+z^2)} \ge 0\]
Therefore
\[\sum \frac{x^5-x^2}{x^5+y^2+z^2} \ge \sum\frac{x^5-x^2}{x^3(x^2+y^2+z^2)} = \frac{1}{x^2+y^2+z^2}\sum{(x^2 - \frac{1}{x})} \ge \newline \ge \frac{1}{x^2+y^2+z^2}\sum{(x^2 - yz)} \ge 0\]

The substitution $(x^2 - \frac{1}{x}$ for $(x^2 - yz)$ it's because of $xyz \ge 1$ and $\sum{(x^2 - yz)} \ge 0$ it's because $\sum{(x^2 - yz)} = x^2 + y^2 + z^2 - xy - yz - xz$ result easy to show by factorizing that it is $\ge 0$

Why not putting the awarded solution to its home topic?

Ha ha! I rushed from the jury meeting to post the cutoffs before conking out in my hotel room (it was rather late!)

Wow. That was fun. I wanna do it again! But I doubt they'd let me. There're already 3+++? people who want to go next year as observer Bs.


calculus
integration
function
complex analysis

[hide="Hint"]
Well the answer is obvious through Poisson's Formula and you could prove it easily with Residue Theory
[/hide]

yep, it is just residue theorem. You have to integrate for example $\frac{z dz}{(z-a)({\rho}^2 - \overline{a}z)}$ over the circle $|z|= \rho$, which is routine computations.

Consider the harmonic function $f(z)=1$. So  according to Poisson's Formula with the function $f$ : \[ \frac 1 {2\pi}\int_{0}^{2\pi}\frac{(\rho^2-|a|^2)d\theta}{|\rho e^{i\theta}-a|^2}=1 \]And we know that $dz=izd\theta$. So $|dz|=|z|d\theta=\rho d\theta$, and easily $\frac1{2\pi} \int_{|z|=\rho}\frac{(\rho^2-|a|^2)|dz|}{\rho|z-a|^2}=1$. Finally we get : \[ \int_{|z|=\rho}\frac{|dz|}{|z-a|^2}=\frac{2\pi\rho}{\rho^2-|a|^2} \]


algebra
polynomial
function
superior algebra
superior algebra solved

Let $f \in \mathbb Z[X]$. For an $n \in \mathbb N$, $n \geq 2$, we define $f_n : \mathbb Z / n \mathbb Z \to \mathbb Z / n \mathbb Z$ through $f_n \left( \widehat x \right) = \widehat{f \left( x \right)}$, for all $x \in \mathbb Z$.

(a) Prove that $f_n$ is well defined.

(b) Find all polynomials $f \in \mathbb Z[X]$ such that for all $n \in \mathbb N$, $n \geq 2$, the function $f_n$ is surjective.

[i]Bogdan Enescu[/i]

a) We want to show that if $n|x-y$ then $f_n(\widehat x)=f_n(\widehat y)$. But $f_n(\widehat x)-f_n(\widehat y)=\widehat{f(x)}-\widehat{f(y)}=\widehat{f(x)-f(y)}=\widehat 0$ because $f(x)-f(y)$ is divisible by $x-y$, thus by $n$.

b) Suppose $\deg f\ge 2$. Then it is clear that we can find an integer $x_0$ so that $M=|f(x_0)-f(0)|>|x_0|$. Now in $\mathbb Z/M\mathbb Z$ we have that $f_M(\widehat0)=\widehat{f(0)}$ and $f_M(\widehat {x_0})=\widehat{f(0)}$ (1) and $\widehat{x_0}\ne \widehat0$ (2). Because $\mathbb Z_M$ is finite in order for $f_M$ to be surjective it must be injective, but the contradiction follows from (1) and (2).

So $\deg f\le 1$. $f(X)=aX+b$. If $a=0$ it is clear that $f_n$ is not surjective for $n\ge 2$. If $|a|>1$ in $\mathbb Z_{|a|}$ we have that $f_n(\widehat x)=\widehat b$ for all $\widehat{x}\in\mathbb Z_{|a|}$, thus $f_n$ is not surjective. In conclusion $f=\pm X+b$ with $b\in\mathbb Z$ arbitrary. It is easy to check that these polynomials satisfy the hypothesis...

[quote="Kuba"]$f_M(\widehat {x_0})=\widehat{f(0)}$ (1) [/quote]

Why?

well $|f(x_0)-f(0)|=M$ So modulo $M$, $f_M(\widehat{x_0})=\widehat{f(x_0)}=\widehat{f(0)}$.


WELL ALL THE BEST FOR PART TEST 3.
HOPE ONE AMONG US IN AOPS IS ALL INDIA TOPPER  :D

thanks,same to u and best of luck to all


If a 3'' by 3'' square is added at each successive stage, what will be the area of the rectangle at Stage 6, in square inches?

[asy]size(250); real textsize = 10pt;
draw(unitsquare);
draw(shift(1.5*right)*unitsquare);
draw(shift(2.5*right)*unitsquare);
draw(shift(4*right)*unitsquare);
draw(shift(5*right)*unitsquare);
draw(shift(6*right)*unitsquare);
label("Stage 1",(.5,0),S,fontsize(textsize));
label("Stage 2",(2.5,0),S,fontsize(textsize));
label("Stage 3",(5.5,0),S,fontsize(textsize));[/asy]

We can see that in Figure 1, there is one square which has $ 3*3\equal{}9$ square inches.

In Figure 2, there are TWO squares, each identical to the square in Figure 1.

In Figure 3, there are THREE squares, each identical to the square in Figure 1.

We can see the pattern: in Figure $ n$, there will be $ n$ squares, each identical to the square in Figure 1.

Thus, in Figure 6, there are 6 squares, each square with 9 square inches.

The total area is $ 9*6\equal{}54 square inches.$

There are $ 9$ square inches added every stage, so stage $ 6$ has $ 6*9\equal{}54$ square inches of area.


Find all functions $f: \mathbb N \to \mathbb N$ so that $ f(xf(y))=yf(x)$ for all $ x,y \in \mathbb N$ .

Put $ x \equal{} 1,y \equal{} t$ in the equation to get $ f(f(t)) \equal{} tf(1)\,\forall t\in\mathbb N\quad(1)$.

Again from the equation we have $ f(yf(x)) \equal{} xf(y)$. Hence $ f(f(xf(y))) \equal{} xf(y)$. Thus $ f(f(t)) \equal{} t$ for $ t \equal{} xf(y)$. Combining this with $ (1)$ we get $ f(1) \equal{} 1$. Thus $ x \equal{} 1,y \equal{} t$ implies $ f(f(t)) \equal{} t\,\forall t\in\mathbb N\quad(2)$.

Assume that $ f(a) \equal{} f(b)$. Then we have $ f(af(a)) \equal{} f(af(b))$ which implies $ af(a) \equal{} bf(a)\implies a \equal{} b$. Thus $ f$ is injective. However, $ (2)$ implies that $ f$ is surjective. Thus $ f$ is bijective.

The substitution $ y \equal{} f(z)$ in our original equation yields $ f(f(z)f(x)) \equal{} xf(f(z)) \equal{} zx\,\forall z,x\in\mathbb N$. But $ zx \equal{} f(f(zx))$. Thus $ f(f(zx)) \equal{} f(f(z)f(x))\implies f(zx) \equal{} f(z)f(x)$. This is Cauchy's equation with solution  $ f(x) \equal{} x^k$ (the other solutions don't count since $ f: \mathbb N\to\mathbb N$). But $ f(f(x)) \equal{} x$ implies $ k^2 \equal{} 1$. Hence $ k \equal{} 1$ (as $ k\neq \minus{}1$). Thus the only solution is $ \boxed{f(x) \equal{} x}$.

[quote="zaya_yc"]Find all function $ f: N\to N$ so that $ f(xf(y)) \equal{} yf(x)$ for all $ x,y \in N$ .[/quote]

Let $ P(x,y)$ be the property $ f(xf(y)) \equal{} yf(x)$
Let $ f(1)\equal{}a$

$ P(1,1)$ implies $ f(a)\equal{}a$
$ P(1,a)$ implies $ a\equal{}a^2$ and so $ a\equal{}1$
$ P(1,x)$ implies $ f(f(x))\equal{}x$ and $ f(x)$ is bijective

$ P(x,f(y))$ implies $ f(xy)\equal{}f(x)f(y)$ and $ f(x)$ is multiplicative and need only to be defined for primes.

$ f(x)\equal{}ab$, with $ a,b>1$ implies then $ x\equal{}f(a)f(b)$ with $ f(a)>1$ and $ f(b)>1$ (since f is bijective and so only $ f(1)\equal{}1$) and so image of a prime is a prime and only primes give primes.

Split then the set of primes in three distinct subsets $ A$, $ B$ and $ C$ such that exists a bijection $ h(p)$ from $ A$ to $ B$.
Let then $ f(x)$ defined as :

$ f(1)\equal{}1$
$ f(p)\equal{}h(p)$ $ \forall p\in A$
$ f(p)\equal{}h^{\minus{}1}(p)$ $ \forall p\in B$
$ f(p)\equal{}p$ $ \forall p\in C$

$ f(\prod p_i^{n_i})\equal{}\prod f(p_i)^{n_i}$ $ \forall$ non-primes.

And it is easy to verify that these necessary solutions are sufficient.

[quote="nayel"]$ f(zx) \equal{} f(z)f(x)$. This is Cauchy's equation with solution  $ f(x) \equal{} x^k$ (the other solutions don't count since $ f: \mathbb N\to\mathbb N$). [/quote]

Hello Nayel !

You're wrong here. Cauchy multiplcative equation has a lot of solutions different from  $ f(x) \equal{} x^k$ when in $ \mathbb{N}$. For example, here :

$ f(1)\equal{}1$
$ f(2)\equal{}3$
$ f(3)\equal{}2$
$ f(p)\equal{}p$ for any prime $ p>3$

And other values using multiplicative property ($ f(4)\equal{}9$, $ f(8)\equal{}27$, $ f(9)\equal{}4$, $ f(12)\equal{}18$, ...).


Hi,

find the greatest constant $ c$ such that

\[ {{y^2z^2}\over{x^2(x\minus{}y)(x\minus{}z)}}\plus{}{{z^2x^2}\over{y^2(y\minus{}z)(y\minus{}x)}}\plus{}{{x^2y^2}\over{z^2(z\minus{}x)(z\minus{}y)}}> c.\]

[quote="ggbg90"]are there any restrictions on x,y,z? otherwise the answer is 0, the inequality is always positive by my work.[/quote]

Sorry! yes $ x,y,z$ are supposed to be postive distinct real numbers :blush:


Khob ye zare moarefi bad nist :

har kasi khast ,khodesho moarefi kone  :D  :D :

Khob az baro bache team iran shoro mikonim:

1.moderatore azizeman :

Omid Hatami:
Ahle esfehan az dabiretane ejehi ,va emsal (2005) ozve team iran dar imo bodan  ,va dar hale hazer ,moaleme bashgahe danesh bajohane javan hastan (mahale bargozariye kelas ha baraye ghabolihaye marhale 2vome opamliyade iran )

2:Nima ahmadi pore anari :

yeki az navabeghe iran ,ozve team riyaziye iran dar sale 2005 ,va ehtemalan (100%) ozve team computer iran ham mishe ,dar dabirestane tiz hoshane tehran dars khondan  va (va khodayish in bashar bahoshe )

3.man,behzad ,amir2,ashegh,jensen ,pooyan _seva,...:

ham ham  madreseyi hastim sale sevome dabirestan

khob hala shoma ? :D  :D  :D  :D

aha rasti ye chiz dige :

darbareye medal ha :

omid hatami: noghreye jahani

ahmadi por: noghreye jahani

refighe man behzad :bronze iran ( :D hanoz  jaye bishraf dare na ? ,bale ke dare  :P )

amirhtlusa:
amir hastam, motassefaneh man too iran namoondam ta too olampiad sherkat konam, dar sene 13 salegi oomadam, USA.
hamin :)

man ham ke dige niazi be moarefi nadaram :D  :D ostad tavakolli goftand. ;)  ;)

salam bache ha,man ham milad hastam,va az inke bache ha inja jam mishan kheili khosh halam.(rasti omidvaram baghie bache ha ham bian :D )

sallam .hamoontori ke lomos_lupin e aziz goft man ashegham.va asheghe hendesam.

vaghean jaye aghaye Davood Vakily roo inja khaly mibinam...

pedram heidari ham yeki az bache haye mokhe roozgare...

number theoryish marekast...

pedram khodeto bishtar moarrefi kon... ;)

Salam bar hame. man kochike shoma  ehsan hastam ,va dar tehran zendegi mikonam va sale sevome dabirestanam  ;) hamin  :P

ali ashegh ali hastam too dabirestan e enerji hastam.

Mokhless shomaa Ramtin hastam. ("less" be farsi bood na be english :lol: )

shagerde hamatoon; mostafa hastam va yeki dosali ke az vatan dooram

[quote="saeid"]Chakere hamatooon saeid sajjadi hastam sale avale Dabirestan.[/quote]
ghorboonet. :D

manam mokhlese hamatoon BaBaK hastam
sale dovome dabirestan az allame tabatabai tehran
baraye computer va riazi mikhoonam(bishtar computer)  :)

[quote="mahsa mokh"]mahsa hastam az khavare miyane :rotfl:[/quote]
wow jeddi migi?

agha ye lotfi bokonido sale tahsilitoono ham begid va madrese tahsil
merC :D

amir/17
United States>>Washington>>Seattle>>Eastlake High School.

shoma mikhayn too USAMO sherkat konin? age hamin torie lotfan begin ke chetori shoma ke irani hassin mitoonin sherkat koni?(nakone raftino tabaiyate oonja roo...) har joor doos darin

bale man mikham too USAMO sherkat konam,
baraye sherkat dar USAMO laazem nis ke Citizen bashid, faghat bayad Green Card dashteh bashid, ke man daram :D

motmaenni?
choon do seta az bacheha inja to canada tooye USAMO sherkat kardan va fekr nakonam ke green card dashtand.

vali khob motmaen nistam  :)

[quote="amirhtlusa"]aha, Canada fargh mikoneh...
hameye daneshamoozan e kandayi mitoonan dar USAMO sherkat konan (agar entekhab beshan), vali, nemitoonan too MOP (tabestoon), va tooye IMO team bashan...
baraye inke too MOP ya IMO team bashan bayad Green Card dashteh bashan.[/quote]

Cheghadr omidvari ke be MOP davaat beshi?
tooye Canada ke fekr nakonam bishtar az 10-12 nafar be urduye amadegy davaat beshan.

[quote="mahsa mokh"]khoobe ke khodamo kaamel moarrefi konam:
mahsa dadar az farzaanegan va 17 saale[/quote]

Farzaanegane koja?

do sal ghabl nafar avval physic Canada aghaye Amir Ali Modir Shanechi boodan ke chand sally tooye shahid hashemi nejad mashhad dars khoonde boodan.
Ishoon talaye IPHO barande shodan :)  :)

[quote="fatehi"]
Cheghadr omidvari ke be MOP davaat beshi?
tooye Canada ke fekr nakonam bishtar az 10-12 nafar be urduye amadegy davaat beshan. [/quote]
ziyad na...
man USAMO baram base :)
are canada bache haye zerangi dareh, vali sathe reghabat to USA balatare,
(albatte fekr konam 2-3 ta az canadayi ha parsal USAMO ro barandeh shode boodan)

az khanoome maryam mikham keh khodesho inja moarrefi koneh. ;)

manam maryam rashin hastam ke emsaal gharare beram avvale dabirestan va yeki do sale daram mikhoonam ke babam kheili komakam kard.hala khodetoon hads bezanin chand salame? :?:
rasti man be nazarie adaad alagheye ziyadi daram vali toosh zaifam :(


Solve the easy equation....
(x+1)(x+2)(x+3)(x+6)=12x^2

Nice and simple method required :lol:

Oh man,both eminem and i got the qustion on the same day,looks like he posted it first ;)

Anyway,some more indigenous solutions are welcome here. :lol:


[i]Τυχαία βρήκα σε ένα αμερικάνικο  μαθηματικό διαγωνισμό το παρακάτω κείμενο , γραμμένο από μια μεγάλη μορφή των μαθηματικών και της διανόησης.Είναι κρίμα που ο κάθε αιώνας γεννάει  ελάχιστους τέτοιους ανθρώπους.Όταν διαβάζεις τις σκέψεις τους, νοιώθεις να περνάει από το μυαλό και το σώμα σου το αεράκι του  Μάη που σε φρεσκάρει , σε αναζωγονεί και σε γεμίζει αρώματα. Απολαύστε το , διαβάζοντας το δυο- τρει φορές απανωτά.Και αν είναι δυνατόν να το μάθετε απέξω.Αξίζει ! : [/i]

 [b]ΥΜΝΟΣ ΣΤΗ ΦΥΣΗ[/b]
[color=blue]
'' The scientist does not study nature because it
is useful; he studies it because he delights(την απολαμβάνει )
in it, and he delights in it because it is
beautiful. If nature were not beautiful, it
would not be worth knowing, and if nature were
not worth knowing, life would not be worth living.''[/color]

                                             [color=red]Henri Poincare[/color]


Putnam
floor function
arithmetic sequence
college contests

Call a subset $ S$ of $ \{1,2,\dots,n\}$ [i]mediocre[/i] if it has the following property: Whenever $ a$ and $ b$ are elements of $ S$ whose average is an integer, that average is also an element of $ S.$ Let $ A(n)$ be the number of mediocre subsets of $ \{1,2,\dots,n\}.$ [For instance, every subset of $ \{1,2,3\}$ except $ \{1,3\}$ is mediocre, so $ A(3)\equal{}7.$] Find all positive integers $ n$ such that $ A(n\plus{}2)\minus{}2A(n\plus{}1)\plus{}A(n)\equal{}1.$

This was probably my favorite.

First, we characterize the mediocre sets. Every mediocre set is either a set with $ 0$ or $ 1$ element, or it's an arithmetic progression with odd common difference. If two adjacent elements differ by an even number, the average must lie between them and the set isn't mediocre. If we have $ a\minus{}d,a,a\plus{}d'$ adjacent elements with $ d\neq d'$ both odd, the average $ a\plus{}\frac12(d'\minus{}d)$ is another element in that interval, and the set isn't selfish. Conversely, it's easy to see why an arithmetic progression with odd difference works; if the average of two elements is an integer, those elements differ by an even number of steps, and their average is the element of the sequence halfway between.

Now, we count them. Let $ s(i)$ be the number of odd divisors of $ i$. Then, for $ k>j$, the number of arithmetic progressions with odd difference starting at $ j$ and ending at $ k$ is $ s(k\minus{}j)$. Sum over all $ j,k$ to get $ A(n)\equal{}n\plus{}1\plus{}\sum_{1\le j<k\le n}s(k\minus{}j)$ (the initial $ n\plus{}1$ counts the mediocre sets with at most $ 1$ element).
Now, take a difference:
$ A(n\plus{}1)\minus{}A(n)\equal{}(n\plus{}2)\minus{}(n\plus{}1)\plus{}\sum_{1\le j<n\plus{}1}s(n\plus{}1\minus{}j)\equal{}1\plus{}\sum_{i\equal{}1}^n s(i)$
Take another difference:
$ A(n\plus{}2)\minus{}A(n\plus{}1)\minus{}A(n\plus{}1)\plus{}A(n)\equal{}\sum_{i\equal{}1}^{n\plus{}1} s(i)\minus{}\sum_{i\equal{}1}^n s(i)\equal{}s(n\plus{}1)$
We want to know when this quantity $ s(n\plus{}1)$ is $ 1$. Every number has $ 1$ as an odd divisor; if it has any other odd divisor, its value of $ s$ is greater than $ 1$. That leaves the numbers with no nontrivial odd factors, which are the powers of $ 2$.

Answer: those $ n$ of the form $ 2^k \minus{} 1$.

In this one I proved that $ A(n\plus{}2)\minus{}2A(n\plus{}1)\plus{}A(n)\equal{}\sum_{d\geq 0}\left(\lfloor \frac{n\plus{}1}{2d\plus{}1}\rfloor\minus{}\lfloor\frac{n}{2d\plus{}1}\rfloor\right)\equal{}s(n\plus{}1)$ just like jmerry did and then stated: "So n+1 must have only one odd factor, $ n\plus{}1\equal{}2^kp$ for some odd prime p."  :roll:

[quote="Kent Merryfield"]No wonder I never figured it out. I was misreading and misinterpreting the definition of mediocre - and got a whole different sequence.[/quote]

Cosign.

Another way to solve this: calculate $ A(n \plus{} 2)$ recursively as follows. If we include one of $ 1$ or $ n$ (but not both), then there are $ A(n \plus{} 1)$ possibilities, if we include neither there are $ A(n)$ possibilities, and define $ B(n \plus{} 2)$ to be the possibilities if we include both. Then $ A(n \plus{} 2) \equal{} 2A(n \plus{} 1) \minus{} A(n) \plus{} B(n \plus{} 2)$ so $ A(n \plus{} 2) \minus{} 2A(n \plus{} 1) \plus{} A(n) \equal{} B(n \plus{} 2)$ so we want to know when $ B(n \plus{} 2) \equal{} 1$.

The whole set always works, so we are looking for solutions including both $ 1$ and $ n$ but not the whole set. If $ n \plus{} 2 \equal{} 2^k \plus{} 1$ then we must include the middle element, so inductively we must include all elements so $ B(n \plus{} 2) \equal{} 1$. Otherwise, it is pretty easy to see that adding in the average elements will terminate before the whole set is included. Thus $ n \equal{} 2^k \minus{} 1$.

EDIT: By the way, if I screwed up the last step (I counted $ B(n\plus{}2)$ using sets that were NOT mediocre instead of sets that were, which trivially gives only $ n\equal{}1$), is there any chance that I'll get more than 2 points? I'm not sure how lenient they are with partial credit.

My approach was quite similar to hypo3400. With a couple of bijections you prove that the only n's that work are those who satisfy the following: the only mediocre subset of {1,2...n+2} that contains 1 and n+2 is the set itself. So if I didn't screw up, I got that n+2 should be $ 1mod(2^k)$ for any power of 2 les than n+2. So obviosly I got $ 2^k\minus{}1$

The first step is to prove that arithmetic progressions with odd periods are the only mediocre subsets. Therefore we can determine a mediocre set by its largest and smallest elements and the period of the arithmetic progression.

Call a mediocre subset of {1,...,n+2} primitive if it contains both 1 and n+2. What we see is that the expression on the left is equal to the number of primitive subsets, since we subtract out the mediocre subsets of {1,...,n+1} and of {2,...,n+2} (both of size A(n+1)) and add back their intersection, i.e. the mediocre subsets of {2,...,n+1} (which has size A(n)). What I think a lot of people will lose points for is not being rigorous in their classification of mediocre sets.


Find a closed expression.
$ \sqrt{x\plus{}y\plus{}z\plus{}2\sqrt{xy\plus{}yz\plus{}zx\plus{}2\sqrt{xy^2z\plus{}yz^2x\plus{}zx^2y\plus{}2\sqrt{\cdots}}}}$

What's the pattern in the symmetric sums?

Regarding the pattern:
Let $ a_{1,i}, a_{2,i}, a_{3,i}$ denote the summands under the $ i$th radical. Then with $ a_{1,1}\equal{}x,a_{2,1}\equal{}y,a_{3,1}\equal{}z$ as initial terms, the following recursion gives the subsequent terms:
$ a_{1,i\plus{}1}\equal{}a_{1,i}\cdot a_{2,i}\\
a_{2,i\plus{}1}\equal{}a_{2,i}\cdot a_{3,i}\\
a_{3,i\plus{}1}\equal{}a_{3,i}\cdot a_{1,i}$.


The sum of two numbers $ x$ and $ y$ is 153, and the value of the fraction $ \frac{x}{y}$ is 0.7. What is the value of $ y \minus{} x$?

assume $ x$ is 7 and $ y$ is 10
then adding you find it's only $ 17$
to make it $ 153$ you need to multiply by 9
to get the real difference we need to multiply by 9 as well
so $ 3\cdot 9\equal{}27$
27.
basically ratio logic without the colons


prove that however we choose n+1 numbers of out the first 2n positive integers we will always obtain two co-prime numbers.

divide the set of the first two 2n positive integers into n sets with two consecutive integers each. by the pigeonhole principle, since we choose n+1 integers, we will have at least one set of two consecutive numbers, which are coprime.

super. now try to prove that among the n+1 chosen numbers you can find two such that one is the divisor of the other one.

This is equivalent to finding 2 numbers a and b so that a|b.

Lets say that all numbers from the set {1,2,..,2n} are in the following form:
k=c_k* 2^(b_k) with c_k odd and we choose n+1 from the set.
Now k<=2n implies that c_k<2n and between 1 and 2n we only have n odd numbers, so at least 2 from them (from the n+1 numbers) have the same c.

so  if c_q=c_w  and b_q>=c_w then q|w.

Nice Lagrangia : You made your first step into problem-solving :D

Do not wait for others 900 messages before trying another... :D

Pierre.


If y=2(4-x)(x+root[(x)^2+196],
find the max. value of y

Very long method...
any short one????(without calculus)

Putting $ x = 14\sinh t$, we get

\begin{eqnarray*}y & = & 2(4 - 14\sinh t)\cdot 14(\sinh t + \cosh t) \\
& = & 28(4 - 14\sinh t)\cdot e^t \\
& = & 28(4e^t - 7e^{2t} + 7) \\
& = & - 28(7e^{2t} - 4e^t - 7)\end{eqnarray*}

The obtained quadratic function in $ e^t$ (the one in the parentheses) yields its minimum for $ e^t = {2\over 7}\iff x = - {45\over 2}$, and it is $ M = - {16 + 4\cdot 49\over 28}$. Therefore the maximum of $ y$ is $ y_M = 16 + 4\cdot 49 = 212$ for the stated value of $ x$.


1.) A student has 10 different books. Of these, 5 are science books, 2 are history books and 3 are math books. He wishes to arrange the books on a shelf in such a way that books of the same type are together. In how many different ways can he arrange the books?

2.) Determine all integer values of $n$ for which the equation $\frac{1}{x-7} + \frac{n}{x-1} = 1$ has two distinct rational solutions

I'll try #2.

[hide]
Solving the equation for x, I got
$x=\frac{(n+9)\pm\sqrt{n^2-10n+49}}{2}$
Looking at the discriminant of that equation, we have $n^2-10n+49>0$.  However, this equation is always true because the coefficient of the quadratic term is positive and its discriminant is less than zero (-96).

So I'd say all integral values but n=0 let that equation have two rational solutions.

$n\neq0$

Correct me if I'm wrong :)
[/hide]

And while I'm at it, I might as well also try #1.

[hide]
Let the string of science books be S, the string of history books be H, and the string of math books be M.
There are 3! ways to arrange the order of the groups on the shelf.

SHM
SMH
HSM
HMS
MSH
MHS

There are 5! ways to arrange the 5 science books.
There are 2! ways to arrange the 2 history books.
There are 3! ways to arrange the 3 math books.

So then I believe the answer would be (3!)(5!)(2!)(3!)=8640.

[/hide]

For #2 the discriminant I got was $n^2 - 10n - 24$ which gives the answer $n>12, n<-2$

Isn't the answer to #1 supposed to be something like 2520?

Nope, the correct answer to number 1 is 8640.

So far, no one has a correct solution to number 2. However, if you want look at the answer (not the solution), take a look and then try to come up with a solution...

[hide]When $n = 4, 6$ or $10$, the given equation will have two distinct rational solutions. This is important, because the problem only wants two distinct and rationals solutions, and nothing else.[/hide]

[hide="number 1"]$5!$ ways to arrange science books, $2!$ ways to arrange history books, $3!$ ways to arrange the math books, and $3!$ ways that the order of the history, science, and math books can be arranged.  So multiplying these all together gives $8640$.[/hide]

[quote="paladin8"][quote="amirhtlusa"]here is #2
$\frac{1}{x-7}+\frac{n}{x-1}=\frac{x-1+nx-7n}{x^2-8x+7}=1\Rightarrow x^2-8x+7=(n+1)x-(7n+1)\Rightarrow x^2-(n+9)x-(7n+8)=0\Rightarrow (n+9)^2-4(7n+8)>0$ since this is true for all $n$, it means $x^2-(n+9)x+(7n+8)=0$ all $n\in\mathbb{R}$ ;)[/quote]

But that doesn't guarantee a [b]rational[/b] solution, does it? We need $(n+9)^2-4(7n+8) = n^2-10n+49$ to be a square and there have to be two distinct solutions...[/quote]
isn't it that for $ax^2+bx+c=0$ to have a rational solution, then $b^2-4ac>0$?

[quote="sr6622"]Can somebody explain #1 again, I was under the impression that you have (10!)/(5!3!2!) = ?
 why just multiply 5!, 3!, 2! and 3!?[/quote]
first treat the three groups of books as one group, you can put them together, in $3!$ ways.
now each group can have different situations
you can put the science books in $5!$ ways, the histroy books in $2!$ and the math books in $3!$ so the total would be: $3!5!3!2!$


vector
superior algebra
superior algebra solved

Let $E$ be an extension field of a finite field $F$, where $F$ has $q$ elements. Let $\alpha\in{E}$ be algebraic over
$F$ of degree $n$. Prove that $F(\alpha)$ has $q^{n}$ elements.

Since $\alpha$ is of degree $n$, the elements $1, \alpha, ..., \alpha^{n-1}$ for a basis of $F(\alpha)$,considered as a vector space over $F$ and every element of $F(\alpha)$ can be written in the form $\sum_{i=0}^{n-1}a_{i}\alpha^{i}, a_{i}\in F$, for every $a_{i}$ we have a choice of $q$ elements.Thus $|F(\alpha)| = q^{n}$.


Not sure how the answer is the way it is but...

Two busses leave Toledo at the same time and travel in opisite directions. One bus averages 55 Miles per hour and the other bus averages 45 miles per hour. After how many hours will the busses be 400 miles apart?

The answer is 13 my teacher said...but just about everyone in class got 4...help me please?

[hide]After 1 hour they are $100$ $miles$ apart. So in $x$ $hours$ it will has to be $400$ $miles$ apart. $400/100=4$.  $4$ $hours$[/hide]

Like I said...the answer is 13. I got 4 but I don't understand why it's 13 (my teacher said it).

Did your teacher do the problem or just looked at the answer key?

Trust me...she did the problem. First she looked at the answer but then she did the problem to make sure it was right. She wanted us to figure it out for homework =p

i tried sovling the problem in a different way to get 13 and i found out that 13 is attained but in a different problem. We have to take notice the avg is 55 miles per hour and other ones AVERAGE is 45 miles.

average does that mean it will travel that distance. To get the average (which is the mean) it is attained from 2 or more trials and then divided by the hours.

I just said a lot of bull crap above and 13 is not possible. Since 13 is prime, only a specific multiplication and division will give 13 but nothing seems to fit it well. (Unless it's asking when will they be 1300 miles apart.

I thought....  since the average is 45 for one possibily it could be
For example 1 mile for the first 12 hours and then 573 miles in the last 13th hour. (To give an average of 45 miles) and the other one is done in a normal scale to make the difference. This would be amibiguous in many ways.

The answer has to be 4 for now but if the teacher asserted it's 13 its probably a different problem.

[quote="i_like_pie"]Even if the car went very slow at first and then very fast to make an average of 45 miles, the total distance and total time will still be the same.

I don't know what your teacher told you, but the answer is 4.[/quote]

Yeah, the answer has to be 4. Your teacher is probably crazy or doesn't know what she's doing. :D

Well, I know she knows what shes doing...but the crazy part...I think that's true =p Anyways, tomorrow I'll find out if she made a mistake or whatever...and I'll post it here =)

How can it be 13? This is SOOO weird. 100 miles/hr. 400 divided by 100=4.... how is it 13?? :p

[quote="Numbies"]Well, we had a test today so she didn't answer the question lol. But, she promised to answer it tomorrow so I'll post it tomorrow.[/quote]

I can't wait. :D

13?? thats wierd...it would help if you could tell us exactly what she did. (her work)

Uhhh...sorry for making all of you sweat but the answer is...4! Apparently, the teacher wanted us to "discuss" the problem and wanted us to get a discussion going about this problem. Well, anyways if you don't understand that, the answer is 4 after all.  :D

she confused the students...
u posted that the students knew it was 4... (everyone got)

at least she could have done like 1~3 but 13 was way out
and easily implied she made a mistake (or in this case made people discuss)

random number..hehe

well from math i know one thing:

if you have an answer that is totally outrangeous compared to a teacher's answer;and that you have double checked your answer mathematically and logically, you are correct.

Well, she knew it was 4...she just wanted us to, I guess you can say, "prove" our answer by drawing a picture or something.

oh i used to HATE that in lower grades
-jorian

i remembered in first grade or something

our teacher said $2+9=20$ and we had to prove it different...it was easier in those days i think  :lol:

[quote="alex__ib"]the answer is 40, the difference between them is 10 so 400:10=40  :P .

Btw you could check for 13 and see it does not work.[/quote]
That's if they're traveling in the same direction.

[quote="alex__ib"]the answer is 40, the difference between them is 10 so 400:10=40  :P .

Btw you could check for 13 and see it does not work.[/quote]

Er, the answer is 4, not 40.

:blush: I am sorry .
well for going in the opposite direction this is fairly easy to check for 4.  :blush:

Again i am so srry for rushing in.

 :blush:


Let $n$ be a positive integer and $M_n$ the set of $1999$ consecutive integers from $n$ to $n+1998$. Let $a_n$ be the number of ways to partition $M_n$ into two nonempty subsets $A$ and $B$ (that is $A \cup B=M_n$ and $A \cap B= \emptyset$) such that the product of all elements in $A$ is equal to the product of all elements in $B$. Evaluate $a_n$ for each positive integer $n$

[hide]For most n, wouldn't it be 0?  In each such set there are primes x such that no other numbers in the set are multiples of x, right?  Like in the one from 1 to 1999, the primes above 1000 are such that, if one of those belongs to A, then the product of the numbers in B will not have that one as a factor.  Therefore it's probably 0...Unless there are really big numbers with intervals of 1000 consecutive integers and no primes in that set such that this works...but there are still other primes which will be factors in those...[/hide]

k81 -- good thoughts.  It'll help to think about [hide]what type of number 1999 is.[/hide]

Related to something you sort-of asked, though, there are arbitrarily long sequences of composite numbers.  If you want $n$ composite numbers, you can always take $(n + 1)! + 2, (n + 1)! + 3, \ldots, (n + 1)! + (n + 1)$.

But there will always be primes p in the set such that in the prime factorization of the product of all the elements of the set, p will only appear once...

Well, yes, you have to prove that :)  It's not to hard -- you just need to find the right p.

Well if we're lucky, 1999 is prime :P.  If it isn't maybe try 1997.  But don't forget that you can have those pesky squares too!

[quote="probability1.01"]Well if we're lucky, 1999 is prime :P.  If it isn't maybe try 1997.[/quote]

Hah, you clearly haven't been doing math contests for long enough.  Every short-question star my age should remember that {1997, 1999, 2003} contains one pair each of [url=http://mathworld.wolfram.com/TwinPrimes.html]twin primes[/url], [url=http://mathworld.wolfram.com/CousinPrimes.html]cousin primes[/url], and [url=http://mathworld.wolfram.com/SexyPrimes.html]sexy primes[/url].  (This is also known as a [url=http://mathworld.wolfram.com/PrimeTriplet.html]prime triplet[/url].)

Well....in that case
If there exists a prime between 1000 and 1999 inclusive, then we're done, and we know that for 1-1999, A and B cannot have the same product.  I say that there exists a prime between 1000 ad 1999.  Who disagrees with me?  We can even just find one by randomly testing...
In order to prove it for a general n, we have to show that there exists at least one prime p in the interval [n, n+1998] such that 2p does now exist in the interval.  We know that if n>1999, then we can't have both p and 2p in the set.  And there exists at least one prime in each of the intervals [n, n+1998] for n=[1, 2000]


For $a,b,c$ positive reals prove that

${\displaystyle \sum \frac{a^2}{2b^2-bc+2c^2} \ge 1}$

[quote]If you solve this challenge you will have like deserved prize Vasc solution for

$\displaystyle \frac{2a^2+qbc}{b^2+pbc+c^2}+\frac{2b^2+qca}{c^2+pca+a^2}+\frac{2c^2+qab}{a^2+pab+b^2} \ge \frac{3(2+q)}{2+p}$  ;)

Good work  :)[/quote]

This is no little. :lol:

\[
\left( {\sum {\frac{{a^2 }}{{2b^2  + 2c^2  - bc}}} } \right)\left( {\sum {a^2 \left( {2b^2  + 2c^2  - bc} \right)} } \right) \ge \left( {\sum {a^2 } } \right)^2
\]
\[
\left( {\sum {\frac{{a^2 }}{{2b^2  + 2c^2  - bc}}} } \right) \ge \frac{{\left( {\sum {a^2 } } \right)^2 }}{{\sum {a^2 \left( {2b^2  + 2c^2  - bc} \right)} }} = \frac{{\left( {\sum {a^2 } } \right)^2 }}{{4\left( {a^2 b^2  + b^2 c^2  + c^2 a^2 } \right) - abc(a + b + c)}}
\]
\[
\begin{array}{l}
 \frac{{\left( {\sum {a^2 } } \right)^2 }}{{4\left( {a^2 b^2  + b^2 c^2  + c^2 a^2 } \right) - abc(a + b + c)}} \ge 1 \\
  \Leftrightarrow a^4  + b^4  + c^4  + abc(a + b + c) \ge 2\left( {a^2 b^2  + b^2 c^2  + c^2 a^2 } \right) \\
 \end{array}
\]
by schur,
\[
a^4  + b^4  + c^4  + abc(a + b + c) \ge \sum\limits_{sym} {a^3 b}
\]
by muirhead,
\[
\sum\limits_{sym} {a^3 b}  \ge 2\left( {a^2 b^2  + b^2 c^2  + c^2 a^2 } \right)
\]
so it is true.
vasc can you post your very nice solution for the very nice ineq???

can we prove $\sum a^{4}+abc\sum a\geq 2\sum a^{2}b^{2} $
I also ask a question:what is shur? I have seen you write it often.but I don't know what it means.
thank you in advance.

[quote="zhaobin"]can we prove $\sum a^{4}+abc\sum a\geq 2%Error. "suma" is a bad command.
^{2}b^{2} $
I also ask a question:what is shur? I have seen you write it often.but I don't know what it means.
thank you in advance.[/quote]

schur refers to this
http://www.mathlinks.ro/Forum/viewtopic.php?t=1166

[quote="Vasc"]I haven't a very nice solution or a nice solution for the very nice inequality (vasc 4 inequality).
I have an ugly solution for $p \ge -1$.[/quote]

maybe our definitions of "ugly" is different....
hehe, so can you post your solution? Thank you very much!!

[quote="Vasc"]I haven't a very nice solution or a nice solution for the very nice inequality (vasc 4 inequality).
I have an ugly solution for $p \ge -1$.[/quote]
is your solution to prove that:
$(2+p)(2\sum a^{2}+q\sum ab)^{2}\geq 3(2+q)(\sum (b^{2}+pbc+c^{2})(2a^{2}+qbc)$
                                                                                :(

Siuhochung, if I post now my ugly solution for $p\ge -1$, who will nicely solve vasc3 ineq, vasc4 ineq, (C) ineq for $p=1$ and (D) ineq for $p=2$ ? :(

Zhaobin, in my solution for $p\ge -1$, I don't use Cauchy-Schwarz ineq (which fails for $p$ positive). ;)

[quote="manlio"]For positive numbers $ a,b,c$ prove that

$ \frac {a^2}{2b^2 - bc + 2c^2} + \frac {b^2}{2c^2 - ca + 2a^2} + \frac {c^2}{2a^2 - ab + 2b^2} \ge 1.$[/quote]${ \sum{\frac {a^2}{2b^2 + 2c^2 - bc }}\geq\sum{\cfrac {a^2}{2(b^2 + c^2) - \cfrac{2b^2c^2}{b^2 + c^2}}}} = \sum{\frac {a^2(b^2 + c^2)}{2(b^4 + c^4 + b^2c^2) }}\geq 1,$

the last inequality is the Vasc0 strange inequality :

http://www.mathlinks.ro/Forum/viewtopic.php?t=25553

in there, a nice and unexpected solution be posted by Vasc.

By the way, we give another direct proof :

$ \frac {x^2}{2y^2 - yz + 2z^2}+\frac {y^2}{2z^2 - zx + 2x^2}+\frac {z^2}{2x^2 - xy + 2y^2}-1$

$ =\sum{\frac{(y-z)^2[4(y+z-x)^2(x^2+y^2+z^2)+xyz(x+y+z)+4yz(y-z)^2]}{2\left(2y^2-yz+2z^2)(2z^2-zx+2x^2)(2x^2-xy+2y^2)}}$

$ \geq 0.$


calculus
integration
calculus computations

What is the indefinite integral of $r*e^{r}$ with respect to $r$

What is the basic formulation of integration by parts? This is, more or less, the very first problem you would do upon learning about integration by parts.

Ah, okay I am an idiot, I got it. This shameful event has inspired me to go back to some calc studies immediately, lol.


Unless, I 've done soething silly, I get a contradiction.
Writing the system as
ab^2 + bc^2 +ca^2 = 23
a^2b^3c + b^2c^3a + c^2a^3b = 150
a^2b^2c^2 = 36,
I get abc = 6 or -6.
Then factorising the secod equation,
abc(ab^2 + bc^2 + ca^2) = 150,
ab^2 + bc^2 + ca^2 = 25 or -25,  and not 23.

Thank you for your pointing out my mystake.

I have just edit it.

kunny

The first and second equation are equivalent, the second is just $ \sqrt {xyz} \equal{} 6$ times the first.
So I think there are infinitely many solutions. But it we want integer solutions, we can just guess and check the factors of 36, we see x=9 y=1 z=4 and permutations work.


Does anyone have a complete AIME 2005 B solutions??

Thanks,


algebra
polynomial
algorithm
number theory
Euclidean algorithm

Solve the system

$ a(b\plus{}c\plus{}d\plus{}e\plus{}f)\equal{} 184$

$ b(a\plus{}c\plus{}d\plus{}e\plus{}f)\equal{}225$

$ c(a\plus{}b\plus{}d\plus{}e\plus{}f)\equal{}301$

$ d(a\plus{}b\plus{}c\plus{}e\plus{}f)\equal{} 369$

$ e(a\plus{}b\plus{}c\plus{}d\plus{}f)\equal{}400$

$ f(a\plus{}b\plus{}c\plus{}d\plus{}e)\equal{}525$

Adding these equations and divided by 2 gives $ ab\plus{}ac\plus{}ad\plus{}ae\plus{}af\plus{}bc\plus{}bd\plus{}be\plus{}bf\plus{}cd\plus{}ce\plus{}cf\plus{}de\plus{}df\plus{}ef\equal{}1002.$

well, yeah, but...does that help  :huh:

does that help in a really nifty elim thing?

One thing that may help is to add a variable: $ s \equal{} a \plus{} b \plus{} c \plus{} d \plus{} e \plus{} f$. Then the system looks like
\[ [a(s \minus{} a),b(s \minus{} b),c(s \minus{} c),d(s \minus{} d),e(s \minus{} e),f(s \minus{} f)] \equal{} [184,225,301,369,400,525]
\]

As a comment, mathematica tells us that there are a lot of ugly solutions to this system. But the nice one is: [hide]a == 4 && b == 5 && c == 7 && d == 9 && e == 10 && f == 15 && s == 50 (and switching the signs)[/hide]

[quote="t0rajir0u"]The Grobner basis of this system under lexicographic order contains a polynomial in $ f$ of degree $ 52$, so I assume you wanted integer solutions.[/quote]

Where do you come up with these random things? haha

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

http://en.wikipedia.org/wiki/Gr%C3%B6bner_basis

so I must ask, how do these things help and how do we calculate that number of 52?

Grobner basis algorithms are how Mathematica solves systems like this.  If you're interested, [url=http://www.amazon.com/Ideals-Varieties-Algorithms-Computational-Undergraduate/dp/0387946802]Cox, Little, and O'Shea[/url] is a great introduction.

Computing a Grobner basis is roughly a combination of the Euclidean algorithm and Gaussian elimination:  starting with a system of polynomial equations, you get an equivalent system that is hopefully simpler.  It is the best-known substitute for the fact that there is no division algorithm for multivariate polynomials in the traditional sense of the word; division with respect to a Grobner basis satisfies a certain useful technical property.

Anyway, the fact that a polynomial of degree $ 52$ is involved suggests the large number of ugly solutions that you said Mathematica gave you.  (Of course, I also used Mathematica to do the computation.)


For any set $ S$ containing real numbers, prove that:

$ \max\{S\}\equal{}\minus{}\min\{\minus{}S\}$ and also $ \min\{S\}\equal{}\minus{}\max\{\minus{}S\}$

Blah...what if $ S$ does not have a maximum? like $ \{x|x<1\}$?

Also what if $ \{x|x\in R\}$.

I am going to assume that $ S$ is bounded above and below.

I think you want the terms "supremium" and "infinium."

Your properties are direct results of writing out the definitions...

$ \text{sup}\, S$ is a number such that $ \text{sup}\, S\ge x$ for all $ x\in S$. Write out the other definition and you get your proof.

Ok I guess I should have been a little more clear. The set $ S$ is a well ordered set of real numbers $ a_{0}, a_{1}, a_{2}, ..., a_{n}$

[hide="here is my solution"]

$ \max\{S\}\equal{} a_{n}$ and $ \min\{S\}\equal{} a_{0}$

$ \minus{}S \equal{}\{\minus{}a_{0},\minus{}a_{1},\minus{}a_{2},\ldots,\minus{}a_{n}\}$

Now make set $ \minus{}S$ well ordered to get:

$ \minus{}S \equal{}\{\minus{}a_{n},\ldots,\minus{}a_{2},\minus{}a_{1},\minus{}a_{0}\}$

$ \max\{\minus{}S\}\equal{}\minus{}a_{0}$

$ \minus{}\max\{\minus{}S\}\equal{} a_{0}\equal{}\min\{S\}$

And the solution for $ \minus{}\min\{\minus{}S\}\equal{}\max\{S\}$ is practically the same.

[/hide]

BTW, I can't find anything on this supremium/infinium function you speak of.

[quote="Prime factorization"][hide="here is my solution"]
$ \max\{S\}\equal{} a_{n}$
$ \vdots$
$ \minus{}\max\{\minus{}S\}\equal{} a_{n}\equal{}\min\{S\}$
[/hide]
[/quote]

There is a typo.

Let $ S$ be a [b]bounded[/b] non-empty subset of $ \mathbb{R}$ then $ \sup S \equal{}\minus{}\inf (\minus{}S)$. (Or you dont need to say it is bounded and consider $ \pm\infty$, but that trivial). (Actually I should have said "then the infinimum exists .... blah blah blah").

[hide="Proof"]
So there is an least upper bound $ s_{0}\in S$. Consider $ \minus{}S$, this set has a lower bound, i.e. $ \minus{}s_{0}$ because since $ s\leq s_{0}\implies\minus{}s_{0}\leq\minus{}s$. Now suppose that $ \minus{}s_{0}$ aint the great lower bound. Say there is $ \minus{}s_{0}\leq\minus{}s_{1}$ with $ s_{1}\in S$ then $ s_{1}\leq s_{0}$ and $ s_{1}$ is an upper bound on $ S$ since $ \minus{}s_{1}\leq\minus{}s\implies s\leq s_{1}$ but then $ s_{0}\equal{}s_{1}$ since $ s_{0}$ is the supremum.
Q.E.D.
[/hide]


hi
is x^x integrable??if yes what is its primitive? :(

Have a look over the following:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=152766


limit
integration
calculus
calculus computations

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=263687]Already posted[/url].

Where is this coming from that several people would be posting the same problem at once?

(And it's not a very satisfying problem - I suspect there's a typo in the limits of integration.)

Since these are new users, it is reasonable to assume they are in the same class.


Let $ a_i,b_i$ $ (i\equal{}1,2,...,n)$ be real numbers such that the $ a_i$ are distinct, and suppose that there is a real number $ \alpha$ such that the product $ (a_i\plus{}b_1)(a_i\plus{}b_2)...(a_i\plus{}b_n)$ is equal to $ \alpha$ for each $ i$. Prove that there is a real number $ \beta$ such that $ (a_1\plus{}b_j)(a_2\plus{}b_j)...(a_n\plus{}b_j)$ is equal to $ \beta$ for each $ j$.

Consider the polynomial $ f(x) \equal{} (x \plus{} b_1)(x \plus{} b_2)(x \plus{} b_3) \cdots (x \plus{} b_n) \minus{} \alpha$.
All $ a_i$ ($ i \equal{} 1,2, \cdots, n$) are roots of $ f(x)$. Thus
$ f(x) \equal{} (x \minus{} a_1)(x \minus{} a_2)(x \minus{} a_3)\cdots (x \minus{} a_n)$.
Also, $ f( \minus{} b_j) \equal{} \minus{} \alpha$ for all $ j \equal{} 1, 2, \cdots, n$.
Thus
$ ( \minus{} b_j \minus{} a_1)( \minus{} b_j \minus{} a_2)( \minus{} b_j \minus{} a_3) \cdots ( \minus{} b_j \minus{} a_n) \equal{} \minus{} \alpha$,
and
$ (a_{1} \plus{} b_{j})(a_{2} \plus{} b_{j})\cdots (a_{n} \plus{} b_{j}) \equal{} ( \minus{} 1)^{n \plus{} 1}\alpha$.
Let $ \beta \equal{} ( \minus{} 1)^{n \plus{} 1}\alpha$ and we are done.