let $ a,b,c\ge 0$ such that $ ab \plus{} bc \plus{} ca \equal{} 1$

prove that

$ \frac {a}{bc \plus{} 1} \plus{} \frac {b}{ac \plus{} 1} \plus{} \frac {c}{ab \plus{} 1}\ge \frac {3\sqrt{3}}{4}$

We have $ \sum \frac{a}{bc\plus{}1} \equal{} \sum a \minus{} abc\sum \frac{1}{bc\plus{}1} \ge \sqrt{3}\minus{}abc\sum \frac{\sqrt[4]{27}}{4\sqrt[4]{bc}}$. Thus it is left to prove
\[ \frac{1}{\sqrt[4]{3}} \ge \sum x^4y^3z^3,\]
where $ x^4y^4\plus{}y^4z^4\plus{}z^4x^4\equal{}1$. (hence $ a\equal{}x^4$ and so on)
This can be easily done by AmGm.

[quote="matheas"]let $ a,b,c\ge 0$ such that $ ab \plus{} bc \plus{} ca \equal{} 1$

prove that

$ \frac {a}{bc \plus{} 1} \plus{} \frac {b}{ac \plus{} 1} \plus{} \frac {c}{ab \plus{} 1}\ge \frac {3\sqrt {3}}{4}$[/quote]

We can get $ p\equal{}a\plus{}b\plus{}c\geq \sqrt{3},r\equal{}abc\leq \frac{1}{3\sqrt{3}}$

\[ \frac {a}{bc \plus{} 1} \plus{} \frac {b}{ac \plus{} 1} \plus{} \frac {c}{ab \plus{} 1}\]
\[ \geq\frac{(a\plus{}b\plus{}c)^2}{3abc\plus{}a\plus{}b\plus{}c}\geq \frac{(a\plus{}b\plus{}c)^2}{\frac{1}{\sqrt{3}}\plus{}a\plus{}b\plus{}c}\]

we need to prove $ 4p^2\geq 3\plus{}3\sqrt{3}p\Longleftrightarrow (p\minus{}\sqrt{3})(4p\plus{}\sqrt{3})\geq 0$
which is obvious .

USe titu's lemma and then apply amgm to the smplified form

@akashram
please post your solution instead of claiming.

For $ a,b,c$ positive reals, $ a\plus{}b\plus{}c\equal{}1$, the following inequality holds:
$ \sum \frac{a}{bc\plus{}1} \ge \frac{9}{10}$.

[quote="Agr_94_Math"]For $ a,b,c$ positive reals, $ a \plus{} b \plus{} c \equal{} 1$, the following inequality holds:
$ \sum \frac {a}{bc \plus{} 1} \ge \frac {9}{10}$.[/quote]
Yes its too easy...

$ LHS \ge \frac {1}{3abc \plus{} 1}$.
So it suffices to prove that:

$ \frac{1}{3abc \plus{} 1} \ge \frac {9}{10} \Longleftrightarrow \frac {1}{27} \ge abc$.
which is true....(just am-gm on the condition).

[quote="Agr_94_Math"]For $ a,b,c$ positive reals, $ a \plus{} b \plus{} c \equal{} 1$, the following inequality holds:
$ \sum \frac {a}{bc \plus{} 1} \ge \frac {9}{10}$.[/quote]
[hide="Alternate Solution"]We have $ \sqrt{bc}\le \frac{b\plus{}c}{2}\equal{}\frac{1\minus{}a}{2}\implies bc\le \frac{(1\minus{}a)^2}{4}$ and cyclic.

Since $ \frac{a}{bc\plus{}1}\plus{}\frac{b}{ca\plus{}1}\plus{}\frac{c}{ab\plus{}1}\ge \frac{a}{\frac{(1\minus{}a)^2}{4}\plus{}1}\plus{}\frac{b}{\frac{(1\minus{}b)^2}{4}\plus{}1}\plus{}\frac{c}{\frac{(1\minus{}c)^2}{4}\plus{}1}$, it suffices to prove that $ f(a)\plus{}f(b)\plus{}f(c)\ge \frac{9}{10}$ where $ f(x)\equal{}\frac{4x}{(1\minus{}x)^2\plus{}4}.$

Since $ f(x)$ is convex in $ (0,1)$, we may apply Jensen's Inequality, and we have $ f(a)\plus{}f(b)\plus{}f(c)\ge 3\left(\frac{a\plus{}b\plus{}c}{3}\right)\equal{}\frac{9}{10}$, as desired. $ \blacksquare$[/hide]

Another proof using the generalized Means Inequality:
Since $ a\plus{}b\plus{}c\equal{}1$, assigning the weights $ a,b,c$ to the expressions $ \frac{1}{bc\plus{}1}, \frac{1}{ca\plus{}1}, \frac{1}{ab\plus{}1}$ respectively, using weighted AM-HM, it is done.
That is $ M_1(a) \ge M_{\minus{}1} (a)$.

[quote="Agr_94_Math"]Another proof using the generalized Means Inequality:
Since $ a \plus{} b \plus{} c \equal{} 1$, assigning the weights $ a,b,c$ to the expressions $ \frac {1}{bc \plus{} 1}, \frac {1}{ca \plus{} 1}, \frac {1}{ab \plus{} 1}$ respectively, using weighted AM-HM, it is done.
 :lol: That is $ M_1(a) \ge M_{ \minus{} 1} (a)$.[/quote] This method is exactly the same as C-S and thus Dimitris X's proof:

For weights $ \sum w_i\equal{}1$, we have $ \sum \frac{w_i}{x_i}\ge\frac1{\sum w_ix_i}\equal{}\frac{(\sum w_i)^2}{\sum w_ix_i}$  :lol:


writing this for a friend:

he has just downloaded TeXShop for mac from the site directed by aops.  he is using my preamble (that works fine on my computer running the same program), but when he attempts to typeset the code, an error message comes up that says "LaTeX Error: File \(.cls' not found, Type X to quit or <RETURN> to proceed, or enter new name. (Default extension:cis).

Anyone know what the problem/a fix is?  thanks.

Are you absolutely sure it says  \(.cls? Get him to compile a very simple document:
[code]\documentclass{article}
\begin{document}
Hello World!
\end{document}[/code]
Does that work OK? If it does, then there's something wrong with the way your preamble is seen on the Mac, which could be to do with different line endings or whatever (assuming you are using a different operating system). Get him to type (not copy and paste) at least the first couple of lines himself on the Mac and see if that works or gives a different error.


Sa se rezolve ecuatia:

                    \[ \log_{3}\left(2^{\log_{3}(2^{x}+1)} +1\right)=\log_{2}\left(3^{x}-1\right).  \]

 [i] Cezar Lupu, RNMO 2006 shortlist [/i]

Mm...mpai f(x)= $\log_{3}\left(2^{x}+1\right)$ este strict crescatoare. Apoi, daca exponentiem cu baza 2, trecem 1 in stanga si extragem logaritm in baza 3 inegalitatea este echivalenta cu f(f(f(x)))=x (pe un domeniu D pe care asta se poate, la care ma voi referi de acum incolo).
 Cum f(x) este strict crescatoare pe D, daca f(x)<x => f(f(x))<f(x)<x=> f(f(f(x)))<f(x)<x, fals. Idem pentru f(x)>x. Deci ramane f(x)=x singura posibilitate, care in mod evident convine ca solutie daca x este in D. Pai f(x)=x ca ecuatie pe $R$ se transcrie ca $2^{x}+1=3^{x}$ => $1+(\frac{1}{2})^{x}= (\frac{3}{2})^{x}$. In stanga avem o functie descrescatoare, in dreapta una crescatoare, deci ele nu se pot intalni decat in maxim un punct. Pentru x =1 avem egalitate, deci x=1 este singura posibilitate si apoi verificand inapoi vedem ca x=1 este solutie.

   Hehe...imi amintesc de kilul de probleme cu logaritmi si exponentiale pe care le-am facut si speram sa se rezolve asa....dar niciodata nu ieseau in felul asta:(


What would happen if scientific notation took over every single number we wrote? For example, my birthday, February 16, would be $ 2 \times 10^0$ $ 16 \times 10^1$. Every single number would be written like that.

What would be your feedback/comment on this?

I would get used to it but it would never happen because it's counter-productive.  When writing, the goal is to take up as little space as possible and writng 2 as $ 2\times 10^{0}$ takes up more space.

Its only useful when dealing with numbers whose absolute values are really big/really small.

To be more precise, scientific notation is really only useful when describing numbers that have dimensions and when describing relative orders of magnitude - for example, the Planck length or Avogadro's number.  The need for scientific notation is an observation about the relative orders of magnitudes of various phenomena that we describe in science, so to apply it to things like calendars is totally nonsensical.

In fact, when working with quantities so big or so small that they require scientific notation, we often just dump the mantissa and pay attention to just the magnitude.

This is part of my math project, and my project was "what if scientific notation replaced all numbers?". I just wanted some comments, which I got some really good ones, thanks.  :) Any more feedback would be greatly appreciated- maybe a quote promoting scientific notation taking over numbers.

this may not be informative and it also may be wrong, but if scientific notation took over everything, then everyone would be smart :D I'll post something more relevant later...

Isn't scientific notation kinda long though?  Like 10^2*6.2, instead of 620?

That's a point I made earlier.  Scientific notation is only useful when describing relative orders of magnitude, as t0rajir0u pointed out.  When comparing the masses or lengths or velocities or whatever of two bodies that greatly differ, we don't care about the mantissa so much, rather just whatever the power of 10 is.

[quote="KMST"]Scientific notation has a more precise and accurate meaning. It is useful to mean what you say and say what you mean. If something is said to be two billion years old, someone may expect it to be 2,000,000,001 yeas old a year later. If it's specified as 2.0X10^9, you have an idea of the precision of that measurement. It also becomes more universally understood than "2 billions," since "2 billions" means 2,000,000,000 in some countries, and 2,000,000,000,000 in others.

I agree with that icy person ([b]icemathematician7[/b]) who said that if we were using scientific notation we would all be smart (or at least very concerned about accuracy and precision of our data). Maybe we would all be using SI units only, and giving our weight in Kilograms rather than pounds or stones. Bye-bye grains, short tons, and metric tons too. Probably we would refer to temperature in Kelvins. How does a chilly 2.78X10^2 Kelvin sound to you? We could forget all about converting between psi, Torr, atmospheres and Pascals for pressure too.

However, [b]D3m0n Shad0w[/b], your birthday would probably be distilled to a single number. For example, when you tell an Excel spreadsheet that you'll have a birthday party on 16-Feb-09, the number it saves is 39860 (the number of days after the 1990 New Year). Of course, it would be 3.9860X10^4 For more precision, you would tell me what time to show up for your party, for example 3.9860625X10^4.[/quote]

I see what you mean. It would be hard to learn it at an early age and identify the concept of it, leaving most children clueless about numbers and might not learn them later in life. It would be more precise, which would be what we need. And adding seconds after the 1990 New Year is a pain >.<

Do you mind if I quote you as an anonymous person?


Prove that $ 30x^4\plus{}300xy\plus{}100y^3>102y\plus{}306xy^2$ whenever $ x,y\geq 1$.

[quote="MeKnowsNothing"]Prove that $ 30x^4 \plus{} 300xy \plus{} 100y^3 > 102y \plus{} 306xy^2$ whenever $ x,y\geq 1$.[/quote]

try $ x\equal{}6,y\equal{}11$.


Hi everyone,

can someone help me for that question, where do I start, I'm doing integration at the moment.

[quote]Write down a definite integral that will give the value of the area under the curve y= x^2 cos(1/5x) between x=1/2π and  x= 5/2π.
(the expression x^2 cos(1/5x) takes no negative values for 1/2π≤x≤5/2π.
You are not asked to evaluate the integral by hand.)[/quote]

Thanks again :lol:

This should be placed in the Calculus forum.

Thanks JRav, I didn't know where to post it, just new on this website :blush:

[quote]Write down a definite integral that will give the value of the area under the curve y= x^2 cos(1/5x) between x=1/2π and  x= 5/2π.
(the expression x^2 cos(1/5x) takes no negative values for 1/2π≤x≤5/2π.
You are not asked to evaluate the integral by hand.)[/quote]

Just write the integral from 1/2n to 5/2n of x^2cos(1/5x)dx, and then do integratioin by parts of x and xcos(1/5x).

I think he wants to actually know how and why to set it up, not just the answer.

Since $ a\equal{}\frac{1}{2n}<\frac{5}{2n}\equal{}b\forall n>0$, you can set $ a$ as your lower limit and $ b$ as your upper limit.  Since the integral is always positive, you need only one integral, instead of having to split it based on its zeroes.  So the integral is just
$ \int_{\frac{1}{2n}}^{\frac{5}{2n}}x^{2}\cos\left(\frac{1}{5x}\right)\ dx$

JRav, I think the limits of integration are $ \frac {1}{2 \pi}$ and $ \frac {5}{2 \pi}$.

Upon closer inspection, I think you're right.  It's kind of hard to tell.  But my integral is still correct.


function
algebra
domain
real analysis
real analysis unsolved

Prove that a function $u(x,y,z)$ which is harmonic on a reasonable domain $D$ in $R^3$ is determined by its values on $\partial D$.

it suffices to prove that if an harmonic function is null on the boundary of its domain, then this is the null function. But an harmonic function achieves its extremums on the boundary ..


geometry
incenter
circumcircle
geometric transformation
homothety
parallelogram
geometry proposed

Let ABC be an triangle with the incenter I. (I) touches BC, CA, AB respectively at D, E, F. Let D', E', F' be points on (I) such that DD', EE', FF' are diameters of (I).
Denote A' the second intersection of DM and (I) where M is the midpoint of EF. Similarly, we have B', C'. Prove that AA', BB', CC' are concurent at K.

Why did you define $D',\ E',\ F'$? I don't think you used them afterwards..

This is a special case of the following general theorem:

If the incircle of a triangle ABC touches the sides BC, CA, AB at D, E and F, and if A', B', C' are three arbitrary points on the incircle such that the lines DA', EB', FC' concur, then the lines AA', BB', CC' concur, too.

In your case, the lines DA', EB', FC' are the medians of the triangle DEF and hence concur. Thus we get your result.

For a proof of the above general theorem, see the note "Variations of the Steinbart Theorem" [url=http://de.geocities.com/darij_grinberg/]on my triangle geometry website[/url].

  Darij

I 'm really sorry when I mistyped it. Let A' be the intersection of D'M and (I).  :(
  My first proof also uses the theory that Darij said. But my latest one doesn't. One can prove that I,O,K are collinear.

You wrote that I, O and K are collinear. In fact, the point K where the lines AA', BB' and CC' concur turns out to be the internal center of similitude of the circumcircle and the incircle of triangle ABC. I have been trying to prove this working with a heavy arsenal from Steinbart theory (*) and projective geometry. But suddenly everything turned out to be much easier than expected!

We will prove that the lines AA', BB' and CC' concur at the internal center of similitude of the circumcircle and the incircle of triangle ABC.

Let H' be the orthocenter of triangle DEF, and let the lines DH', EH', FH' meet the incircle of triangle ABC at the points A", B", C" (apart from D, E, F). Then it's pretty easy to show that the triangles ABC and A"B"C" are homothetic. The homothety mapping triangle ABC to triangle A"B"C" has a positive stretch factor; the center T of this homothety lies on the lines AA", BB" and CC". But the homothety mapping triangle ABC to triangle A"B"C" must also map the circumcircle of triangle ABC to the circumcircle of triangle A"B"C" (i. e., to the incircle of triangle ABC). Hence, the center T of this homothety is the internal center of similitude of the circumcircle and the incircle of triangle ABC.

Now it only remains to show that the points A, A' and A" are collinear; once this will be shown, it will follow that the line AA' coincides with the line AA", and hence passes through T, i. e. through the internal center of similitude of the circumcircle and the incircle of triangle ABC; similar statements will hold for BB' and CC', and we will be done.

Since DD' is a diameter of the incircle, we have < DED' = 90, so that $DE\perp ED^{\prime }$. Together with $FH^{\prime }\perp DE$, this becomes FH' || ED'. Analogously, EH' || FD'. Hence, the quadrilateral EH'FD' is a parallelogram, and its diagonals bisect each other. This means that the diagonal D'H' passes through the midpoint M of EF. But the point A' is defined to lie on the line D'M. Hence, the point A' lies on the line D'H'.

Since the points E, F, D', A' are concyclic, $A^{\prime }M\cdot D^{\prime }M=EM\cdot FM$. On the other hand, since < AEI = 90 and < AFI = 90, the points A, I, E, F are concyclic, so that $EM\cdot FM=AM\cdot IM$ (the point M lies on the line AI because the points E and F are symmetric with respect to AI, and M is the midpoint of EF). Hence, $A^{\prime }M\cdot D^{\prime }M=AM\cdot IM$, and the points A', D', A, I are concyclic. Thus, with directed angles modulo 180, we have

  < AA'D' = < AID' = < (AI; DI).

On the other hand, since the points D, D', A', A" are concyclic,

  < A"A'D' = < A"DD' = < (DH'; DI).

Since the lines AI and DH' are both perpendicular to EF (the line AI since E and F are symmetric with respect to EF, and the line DH' as an altitude), we have AI || DH' and thus < (AI; DI) = < (DH'; DI), so that < AA'D' = < A"A'D', and the points A, A' and A" are collinear, qed..

Treegoner, was your proof similar to this?

(*) Here is a lemma I wrote up since I thought it could be useful for the proof, but it turned out not to be necessary:

[b]Lemma.[/b] If B' and B" are two points dividing the side CA of a triangle ABC harmonically, and the lines BB' and BB" meet the circumcircle of triangle ABC at the points Y' and Y" (apart from B), and the tangents to the circumcircle of triangle ABC at C and at A meet at the point Y (so that Y is the B-vertex of the tangential triangle of triangle ABC), then the points Y, Y' and Y" are collinear.

[i]Proof.[/i] If the lines CY' and AY' meet the lines AB and BC at C' and A', respectively, then the Ceva theorem yields

[tex]\frac{AB^{\prime }}{B^{\prime }C}\cdot \frac{CA^{\prime }}{A^{\prime }B}\cdot \frac{BC^{\prime }}{C^{\prime }A}=1[/tex].

On the other hand, since B' and B" divide the segment CA harmonically, we get

[tex]\frac{AB^{\prime \prime }}{B^{\prime \prime }C}=-\frac{AB^{\prime }}{B^{\prime }C}[/tex],

so that

[tex]\frac{AB^{\prime \prime }}{B^{\prime \prime }C}\cdot \frac{CA^{\prime }}{A^{\prime }B}\cdot \frac{BC^{\prime }}{C^{\prime }A}=-1[/tex].

Hence, after Menelaos, the points A', B", C' are collinear. In other words, the points AY' /\ BC, BY" /\ CA, CY' /\ AB are collinear, where the sign /\ stands for "intersection of two lines". Now, Theorem 1 of [url=http://de.geocities.com/darij_grinberg/]my note "Variations of the Steinbart Theorem" cited above[/url] reads:

Let XYZ be the tangential triangle of a triangle ABC, and let $X_1$, $Y_1$, $Z_1$ be three points on the circumcircle of triangle ABC. Then, the lines $XX_1$, $YY_1$, $ZZ_1$ concur if and only if either the lines $AX_1$, $BY_1$, $CZ_1$ concur, or the points $AX_1$ /\ BC, $BY_1$ /\ CA, $CZ_1$ /\ AB are collinear.

In our case, if we set $X_1$ = Y', $Y_1$ = Y" and $Z_1$ = Y', then the points $AX_1$ /\ BC, $BY_1$ /\ CA, $CZ_1$ /\ AB are collinear. Hence, the above theorem yields that the lines $XX_1$, $YY_1$, $ZZ_1$ concur, i. e. the lines XY', YY", ZY' concur. But since XY' and ZY' intersect at Y', we see that the points Y, Y' and Y" are collinear, qed..

This lemma helps proving other theorems about the tangential triangle but it isn't helpful in Treegoner's above problem.

  Darij

Sorry, Darij. It 's such a long time I didn't reply it.
   When I first solved it, my solution is very long. But then I finally find a much more simple than it.
   My solution is similar to yours but different in proving A, A'', A' are collinear.
 It 's sufficient to prove that A''EA'F are cyclic harmonic quadrilateral. We have [D'EA']=[D'FA']  (where [ABC] denotes it's area.) since D'A' passes through the midpoint of EF => D'F.FA' = D'E.EA'.
   On the other hand, we have D'E=A''F, D'F=A''E. Thus we obtain A''F.EA'=A''E.FA', which implies A''EA'F a cyclic harmonic quadrilateral. :cool:
    Anyway, thanks for your beautiful solution, Darij.


inequalities
number theory
greatest common divisor

solve this equation and find all answers in natural numbers:
$x^3+3=4y(y+1)$

[quote="jensen"]we are alive to love each other and we love each other to be alive. [/quote]

jensen
I liked very much your signature

this is probably the meaning of life :)

[hide]Subtract 3 and factor to get $x^3=(2y+3)(2y-1)$. The $\gcd (2y+3,2y-1)$ divides the difference, 4; since each is odd the gcd must be 1. Then each factor is a perfect cube. Let $2y+3=a^3$, $2y-1=b^3$ so that $a^3-b^3=4$. $y$ a natural number $\Rightarrow 2y+3\geq 5\Rightarrow a\geq 2$. $a^3-(a-1)^3=3a(a-1)+1\geq 7>4$, so $a^3-(a-1)^3>a^3-b^3\Rightarrow b^3>(a-1)^3\Rightarrow b>a-1$. But for $a^3-b^3>0$, $b<a$. Then $b$ must be between 2 consecutive integers, a contradiction. Therefore, [b]no solution[/b].[/hide]

[quote="varun"][quote="scorpius119"]
The $\gcd (2y+3,2y-1)$ divides the difference, 4[/quote]

Why is it so?[/quote]

let $d=$ $\gcd (2y+3,2y-1)$
then $\rm d|(2y+3)$  and $\rm 7|(2y-1)$
$\Rightarrow$ $\rm d|(2y+3)-(2y-1)$
$\rm d|4$
i think it's clear now varun  ;)

no it was not so obvious but it was easy to understand it.

Yes, there is notheing so obvious. Otherwise maths will be very easy. :rotfl:


[quote="José"]Consider a circle with centre $O$, and $3$ points on it, $A,B$ and $C$, such that $\angle {AOB}< \angle {BOC}$. Let $D$ be the midpoint on the arc $AC$ that contains the point $B$. Consider a point $K$ on $BC$ such that $DK \perp BC$. Prove that $AB+BK=KC$.[/quote]

1) Let $DK$ cut the circle again at $E$, $DO$ cut the circle again at $F$. $\angle FED = 90$ hence $FE \parallel BC$
2) $EB = FC = FA$ hence $EA \parallel FB$, $AB = EF$
3) Draw parallel from $B$ to $DK$. Let $G$ be the point where it intersects $EF$, then $GE = BK$.
4) Triangles $BFG$ and $CKE$ have equal angles and $BF = EC$ hence $\triangle BFG = \triangle CKE$ hence $FG = CK$ and from 2) and 3) $AB + BK = CK$

Daniel


A sequence $ (u_n)$ is defined by $ (1 \plus{} \frac {1}{n})^{n \plus{} x_n} \equal{} e$. Find the limit of the sequence $ (u_n)$

I'm not sure if this is a valid solution, but here it is anyway.

[hide]Solve for $ x_n$:

$ (1 \plus{} \frac {1}{n})^{n \plus{} x_n} \equal{} e$

$ (n \plus{} x_n)\cdot \ln(1 \plus{} \frac {1}{n}) \equal{} 1$

$ x_n \equal{} \dfrac{1}{\ln(1 \plus{} \frac {1}{n})} \minus{} n$

Note that: $ \lim_{n \to \infty}(1 \plus{} \frac {1}{n}) \equal{} 1$

Thus, using the taylor series approximation $ \ln(1 \plus{} x) \approx x \minus{} \frac {x^2}{2}$:

$ \lim_{n \to \infty} \ln(1 \plus{} \frac {1}{n}) \equal{} \lim_{n \to \infty} \left[\frac {1}{n} \minus{} \frac {1}{2n^2} \right] \equal{} \lim_{n \to \infty}\frac {2n \minus{} 1}{2n^2}$

So, $ \lim_{n \to \infty}x_n \equal{} \lim_{n \to \infty} \left[\frac {2n^2}{2n \minus{} 1} \minus{} n \right] \equal{} \lim_{n \to \infty}\frac {n}{2n \minus{} 1} \equal{} \frac {1}{2}$[/hide]


blogs
algebra
functional equation
number theory

The first problem of the second season is K11. Now, it is posted on the blog at

[quote]  [url]http://projectpen.wordpress.com[/url][/quote]

It is suggested by Alexander Remorov from Canada. Next week, we will meet three solutions for the functional equation!

Hi, all. Problems in Elementary Number Theory 1(2008) No. 1 is now weblished on the blog!

[quote]  [url]http://projectpen.wordpress.com[/url][/quote]


Let $ f(x,y,z)\equal{}x^3y\plus{}y^3z\plus{}z^3x$.
Prove that $ f(x,y,z)\equal{}f(ax\plus{}by\plus{}cz,dx\plus{}ey\plus{}fz,gx\plus{}hy\plus{}iz)$ holds for at least $ 168$ pairs of $ (a,b,c,d,e,f,g,h,i)$.
(Actually, it is also possible to prove that the number of such pairs is less than $ 168$.)

How can you prove both at the same time? :huh:


These are really easy. What's the time limit?

@ humblemastermind - 50 minutes / 20 problems + tie breaker. I'm trying to remember what the tie breaker was.

@ Walk Around The River - 2, 5, 14 are incorrect.

[quote="Walk Around The River"]i did these really fast, so i hope theyre right.

[hide="solutions"]1.  20%=1/5.  360/5=$\boxed{72}$
2. 7-6+5=6.  12-6=$\boxed{6}$
3. 40*3/2=60. 60/3=20+6=$\boxed{26}$
4. 1/2x=3.  x=6.  4/6=$\boxed{\frac{2}{3}}$
5. 72-12=60/2=$\boxed{30}$
6. 11+13+17+19=60=$2^{2}*3*\boxed{5}$
7. 25*2=50.  50-25-10-5=$\boxed{10}$
8. 15 cats, 5 dogs, 3 elephants: 25-23=$\boxed{2}$
9. 24/6=4 $\sqrt{4}=2. 2^{3}=\boxed{8}$
10. 214/5=remainder 4, $\boxed{black triangle}$
11. dean 9 ppg.  x/18=9.5. x=$\boxed{171}$
12. 2 p=1 c, 3 c=1 r.  so 6p=1r.  so 1 calc=$\boxed{600}$
13. 6/3=2.  2*3=6. 6*3=$\boxed{18}$
14. 3, 7=$\boxed{2}$
15. 231213=$\boxed{213}$
16. 6/1+5/2+4/3=$\boxed{\frac{59}{6}}$
17. inside angle 116.  2x=64.  x=$\boxed{32}$
18. similar triangles.  1-2=4-6.  so white triangle has area of 6*3/2=9.  2+16=18-9=$\boxed{9}$ or you can just notice that the two parts are equal.
19. 7, 9, 3, 1.  62/4=r2, $\boxed{9}$
20. $2\frac{n(n+1)}{2}<3000$ so $n(n+1)<3000$ so $n=\boxed{54}$[/hide][/quote]

On the real test I did the same thing as you. For #2 I read twentieth as twelfth.

Tiebreaker: There are two prime numbers between 100-199 such that their tens and units digits are primes and if you take away the hundreds digit the remaining two digit number is a prime. What are these numbers and their sum??

The three-digit numbers, two digit numbers (ten's and one's digits) and the ten's and one's digits (alone) must all be prime.

[hide="Tie Breaker Solution"]This is exactly what I wrote:

The ten's digit can be 2, 3, 5, or 7. The one's digit can only be 3 or 7. 123, 153, and 177 are all divisible by 3. The numbers formed by the ten's and one's digits in 127, 133, and 157 aren't prime. The only two remaining numbers are 137 and 173, whose sum is 310.[/hide]

[quote="laughinghead505"]Wow, those are pretty easy! I think I got them all right :lol:

The solutions that haven't been posted yet:

[hide="2"]$20-5+6-7=\boxed{14}$[/hide]

[hide="5"]$(72-12)/12=\boxed5$[/hide]

[hide="14"]$2*2-1=3; 3*2-1=5; 7*2-1=13; \boxed3$[/hide][/quote]

Yup correct, but as I said earlier on some other thread,the questions for the Senior division harder.

Junior - 7th, 8th, 9th grades.

Senior - 10th, 11th, 12th grades.

Unfortunately, I don't have a copy of the senior test, but vishalarul might.

One of the regional competitions around here.

We don't have regional competitions. :dry:


Hi

Can someone help me with this:

Show that $ 4^{2n}\minus{}1$ is divisible by 15 and all integers $ n\ge1$.

Thanks[/quote]

$ 4^{2}n\minus{}1 \equal{} 16^{n}\minus{}1$

Take this $ \mod 15$, and the $ 16^{n}\minus{}1$ becomes $ 1^{n}\minus{}1 \equal{} 0$

When a number is $ 0\mod m$, it means $ m$ divides it. And that's it.

[hide]use mods.

$ 4^{2n}$ is equal to $ (4^{2})^{n}\equal{}16^{n}$

$ 16\equiv 1\bmod{15}$, so you can simplify this to: $ 1^{n}\minus{}1 (mod 15)$ for any #, $ 1^{n}\equal{}1$ so we can say that since this equation becomes 1-1=0 mod 15, all #s for the equation $ 16^{2n}\minus{}1$ for all #s $ \ge 1$ are diviible by 15.[/hide]


Let $ a,b,c$ be positive real numbers. Prove that
\[ \frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{ab+bc+ca}{a^{2}+b^{2}+c^{2}}\ge 4 \]

I think it's old  :)
Solution1:Apply Cauchy-Schwarz's Inequality:
$ \ LHS\geq \frac{(a+b+c)^{2}}{ab+bc+ca}+\frac{ab+bc+ca}{a^{2}+b^{2}+c^{2}}\geq 4$
Or:
$ \frac{a^{2}+b^{2}+c^{2}}{ab+bc+ca}+\frac{ab+bc+ca}{a^{2}+b^{2}+c^{2}}\geq 2$(AM-GM)

Solution2:Assume $ \ c=max(a,b,c)$
We have:
$ \ M(a-b)^{2}+N(a-c)(b-c) \geq 0$
with:
$ \ M=\frac{1}{ab}-\frac{1}{a^{2}+b^{2}+c^{2}}$

$ \ N=\frac{1}{ac}-\frac{1}{a^{2}+b^{2}+c^{2}}$

Solution3:We have:
$ \frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq \frac{9x}{x+2y}$
with:
$ \ x=a^{2}+b^{2}+c^{2}$
$ \ y=ab+bc+ca$
So,it suffices to prove that:
$ \frac{9x}{x+2y}+\frac{y}{x}\geq 4$
Or:
$ \ (5x-2y)(x-y)\geq 0$ is clearly true with $ \ x\geq y$


What is the value of the expression
\[ \frac {1}{\log_{2}100!} \plus{} \frac {1}{\log_{3}100!} \plus{} \frac {1}{\log_{4}100!} \plus{} \cdots \plus{} \frac {1}{\log_{100}100!}
\]
$ \textbf{(A)}\ 0.01 \qquad \textbf{(B)}\ 0.1 \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ 2 \qquad \textbf{(E)}\ 10$

Thanks a lot!

using the change of base formula to rewrite each fraction as $ log_{100!}{n}$ and then applying the property that $ log b \plus{} log a \equal{} log ab$ we get $ log_{100!}{100!} \implies \boxed{C}$

[quote="cognos599"][hide="solution"]by our handy base change formulae, we can see that:

$ \frac {1}{\log_{n}100!} \equal{} \frac {\log_{n}n}{\log_n100!} \equal{} \log_{100!}n$

applying this throughout, the sum becomes

$ \displaystyle\sum_{k \equal{} 2}^{100}\log_{100!}k \equal{} \log_{100!}100! \equal{} \boxed{1}$[/hide][/quote]

How did you get from $ \frac {\log_{n}n}{\log_n100!}$ to $ \log_{100!}n$.

Also, why does $ \displaystyle\sum_{k \equal{} 2}^{100}\log_{100!}k \equal{} \log_{100!}100!$.

Thanks!

[hide="Initial Thoughts"]When I first looked at this problem, I see that there are two things to deal with, a sum of fractions, and differently based logarithms.  We need a way to deal with these problems.[/hide]
[hide="Solution"]It turns out that these problems are actually related.
We know that $ \frac{1}{\log_{a}b}\equal{}\log_{b}a$ and by this we obtain the equation
$ \log_{100!}2\plus{}\log_{100!}3\plus{}...\plus{}\log_{100!}100$
And we know the sum of logarithms is logarithm of their product so
$ \log_{100!}2\plus{}\log_{100!}3\plus{}...\plus{}\log_{100!}100\equal{}\log_{100!}2*3*...*100\equal{}\log_{100!}100!\equal{}1$
$ \boxed{C}$[/hide]

Use Search function or click Resources (or Contests) on the top right to find your problem. Locked.

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=403808#403808

This is the direct link to the resource page's problem.


Let $ \{x_{n}\}$ is a positive sequence .For any $ n\in\mathbb{N^{\plus{}}}$ we denote $ k_{n}$ is smallest positve integer so that :

$ \sum_{i \equal{} 1}^{k_{n}}\frac{1}{i}\ge\sum_{j \equal{} 1}^{n}x_{j}$

Prove that : $ \lim\frac{k_{n\plus{}1}}{k_{n}}$ exists if only if $ \lim x_{n}$ exists. :)

Please solve it. :( I think it is hard problem. Can you post some hint to solve it if you think it is a easy problem??


Question : 10 x 3 rectangle (float)
A clown walks around the edge of the float...

Float goes at 1m/s
Clown moves at 2m/s (clockwise) around the float...

What is the general equation?

My Work Done:

I worked out the angles and the distances around the float but I need a
general equation.

its a piece wise function so split it up into parts.  it should be easy after that


Second one :

[hide]I find (4,2) as only solution.

y must divide x : x = z*y

then y(z-1)=z, so z-1|z but (z,z-1)=1 so z=2 and y=2.[/hide]

First one:

[color=white]xy - 4x - 3y = -3
yz - 5y - 4z = -5
xz - 5x - 3z = 0

(x-3)(y-4)=9
(y-4)(z-5)=15
(x-3)(z-5)=15

From (y-4)(z-5)=15
(x-3)(z-5)=15 you know that x-3=y-4

Because (x-3)(y-4)=9, therefore, x-3 and y-4 could either be 3 or -3.  Since the restrictions state x and y cannot be negative so x-3 and y-4 must be equal to 3.

Solution: x=6 y=7, z=10[/color]

Nice solution for the second problem!

When I did it I did this:
x-y = x/y
xy - y :^2: = x
xy - x = y :^2:
xy - x - y + 1 = y :^2: - y + 1
(x - 1)(y - 1) = y :^2: - y + 1
x - 1 = (y :^2: - y + 1)/(y-1) = y + 1/(y-1)
y-1 has to divide 1 but since y is positive, y = 2 and then x = 4.

For the first problem, Oasis you are correct also.  The solution seems a little different from what I did but it still works, did you ever think of using symmetry for the problem, the system of equations is symmetric.


Are we supposed to solve it without using a calculator? (I assume the answer will be : "Yes", but..  :blush: )

[hide]THe answer is 94509 with a calculator  :D [/hide]

:rotfl: ok, you've used a calculator.. :)

[hide]i used a caculator and the answer is: 94509[/hide]

Well, ok.. but, anirudh, is there a way to solve it without using a calculator?

Hmm.. I can't see how your contest is related with this problem lol?
Or maybe you just want me to participate in it?

[quote="anirudh"]Look here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=46810[/quote]
do you want us to join the contest or look in the problems???????

[quote="anirudh"][quote="236factorial"]anirudh, you can't really add numbers with exponents the way you do with exponents. The answer isn't $9^7$. this problem looks a lot like one of your contest problems.[/quote]

I know, but click on this link:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=46810]
And for math 92's answer, both[/quote]

why? does it have a discussion there?

[quote="anirudh"][quote="236factorial"]anirudh, you can't really add numbers with exponents the way you do with exponents. The answer isn't $9^7$. this problem looks a lot like one of your contest problems.[/quote]

I know, but click on this link:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=46810]
And for math 92's answer, both[/quote]

i still don't understand why you gave a link there.

Same goes for me..
You still haven't explained if your question was solveable without a calculator..

It can be solved without a calculator, just a lot of work.

there is no point in the link and calculators do the trick. :)

:rotfl: yeah..
Anyway, I'd be glad to see your solution 236! (to solve it without a calc.)

Wow.. looong lol
Anyway, thanks for pointing that out, but, it shows us the calc. is a great tool for those nasty things. :)

is there a way to do this without multiplying it all out?

Sorry guys, I posted the problem wrong.  :oops_sign:

:rotfl: All this topic for nothing...
Well, at least, post the good one  :P

Firstly, you guys need to chill out, this is an exercise not a problem.

Secondly, he changed the "problem" :D

$5^7 \cdot 4^7=20^{7}=(2\cdot10)^7=2^7\cdot2^{10}=\boxed{1280000000}$

Wow, G-UNIt is online obviously  :D

[hide]In exponental form, we can see that the exponents are the same, so we multiply the bases together, to get $20^7$.

And it will come out to G-UNIT's answer.  :D [/hide]

[hide][quote="anirudh"]What is $5^7\cdot4^7$[/quote]

If the exponents are the same, we multiply the bases together:

$5 \times 4 = 20$

The exponents stay the same.

So it is $20^7$, or $1280000000$[/hide]

good job. now $5\sqrt{32}\cdot4\sqrt{4}$

this one is easier, and i think i didn't mess up.  :)  :)  :)  :)  :)  :)  :)  :)  :)  :)  :)  :)  :)

[quote="236factorial"][hide]just multiply to get $20\sqrt128=160\sqrt2$[/hide][/quote]

[quote="anirudh"]good job. now $5\sqrt{32}\cdot4\sqrt{4}$ [/quote]

well, what i did was take make 32 equal 16x2. and take out the 16 on this to get 20 root 2 then multiply it by 8 to get 160root2

That's the alternative way, other than the calculator.

we multiply $4*\sqrt{4}=8$ and multiply that by $20\sqrt{2}$ to get $160\sqrt{2}$ :idea: our answer

\[ 5^7 \times 4^7 =2^7 \times 2^7 \times 5^7 =10^7 \times 2^7 =10^7 \times 128 \] (or if you want only one number  here it is :$1280000000$)


complex analysis
complex numbers
complex analysis unsolved

here's a problem from my complex analysis course:

[b]Let the square root be defined with respect to the cut along the negative imaginary axis {[i]yi: y[/i] ≤ 0}. Determine the complex numbers √(-2), √(-3), and their product √(-2)* √(-3).[/b]

and here are my ideas :idea: :

if √(-2) can be written as [i]i[/i]√(2) and √(-3) can be written as [i]i[/i]√(3). their product, though, can be written one of two ways. the first, √(-2)* √(-3) = [i]i[/i]√(2)*[i]i[/i]√(3) = [i]i*i[/i]*√(2)* √(3) = -1 * √(6) =[u][b] - √(6). [/b][/u] the second, √(-2)* √(-3) = √((-2)*(-3)) = [u][b]√(6)[/b][/u].

so, in one case i get - √(6) and in the other i get + √(6).

 :?: How do I deal with this?
 :?: What all is involved in "determining" a complex number?

In the first case you took a positive root and in the second case you took a negative root.


I'm very confused with my hw... please help!

1. Although the situation of coin flips involves the binomial distribution, the analysis can be approximated by the normal distribution.

a. explain what it means to say that the binomial distribution can be approximated by the normal distribution.

b. use the normal distribution approximateion to find out how likely it would be for a fair coin to give results as unbalanced as 573 heads and 427 tails.

2. Based on Roberto's experience with his brother's coin, give 95% and 99% confidence intervals for the true probability of getting heads with that coin.

3. Do you believe that this was a fair coin? That is, do you think the apparenet preference for heasds is because the coin is not balanced, or is it simply a coincidence? Explain your reasoning

$1.$ $a$.
[hide]The binomial distribution is when we know how many times where going to do something (ie. flipping a coin) and we want to know the probability of getting $x$ results or more.  For example, say we're going to flip a coin 500 times and we want to know whats the probability of getting 10 or less heads.. we can figure this out using the binomial distribution.

The normal distribution on the other hand is based off of theoretically what the outcome should approximately be or something along those lines.  So it won't be exact but it will be pretty darn close as $n$ increases.[/hide]

$1.$ $b.$
[hide]So we have $573$ and $427$ tails.

Based off of the normal distribution of flipping a coin: $p_{0}=.5$ and $\sigma = \sqrt{\frac{p \cdot q}{n}}= \sqrt{\frac{(.5)(.5)}{1000}}=\frac{\sqrt{10}}{200}$.  And we have $573$ heads so the proportion of heads is $p=\frac{573}{1000}$.  Thus $z=\frac{p-p_{0}}{\sigma}=\frac{\frac{573}{1000}-.5}{\frac{\sqrt{10}}{200}}\approx 4.616925384$

So now we use the $normalcdf($ key on the calculator and type in $normalcdf(4.6169, \infty) \approx \boxed{.00000194939 \text{ or }.000194939 \%}$.

Note that for $\infty$ you just type in a bunch of $9$s.[/hide]

$2.$
[hide]So we know he has $573$ heads and $n=1000$.  So $p=\frac{573}{1000}$.  A $95 \%$ confidence has a $z^{*}$ of $1.645$ and for $99 \%$ it is $2.330$.

$95 \% \text{ Confidence Interval: }$

$p \pm z^{*}\sqrt{\frac{p \cdot q}{n}}$
$\frac{573}{1000}\pm 1.645 \sqrt{\frac{\frac{573}{1000}\left( 1-\frac{573}{1000}\right) }{1000}}$;  And simplify to interval form:
$\boxed{(.54727, .59873)}$

$99 \% \text{ Confidence Interval: }$

$p \pm z^{*}\sqrt{\frac{p \cdot q}{n}}$
$\frac{573}{1000}\pm 2.330 \sqrt{\frac{\frac{573}{1000}\left( 1-\frac{573}{1000}\right) }{1000}}$;  And simplify to interval form:
$\boxed{(.53655, .60945)}$[/hide]

$3.$
1st Way - Confidence Interval
[hide]Because $.5$ is not in the confidence interval it is highly unlikely these results occurred by chance.[/hide]

2nd Way - Test
[hide]This is where you'd do a test.  But we technically already did one in part $1.$ $b.$ and the value we calculated was a $p-value$ which was $.00000194939$.  All you gotta do is state hypotheses and a conclusion as follows really:

$H_{0}: p_{0}= .5$
$H_{A}: p_{0}> .5$

Then do the test right here that we did in part $1.$ $b.$ then state the following:

With $p-value=.00000194939$ and an $\alpha=.00001$ we reject the null hypothesis.  Because the probability of this occurring in samples is very low, this provides strong evidence that the coin is most likely unfair.

Something along those lines.[/hide]

thanks for the explanations! it is going to take me a while to read through and understand them   :D.


[b]JMC: Level One: Round One: Questions[/b]
This first round has ten questions.  A correct answer to each question is worth two points.  A student may send in solutions to each question for an extra point.  The extra point will be awarded to solutions that show an extensive amount of creativity.  Calculators are permitted for this round, although they are not necessary.  The time given for this round is one week.

http://yashar.biz/jmc/jmc_files/JMC%20Level%20One%20Round%20One%20Questions.pdf

[b]
Questions, comments, and solutions should be e-mailed to jiftedland at gmail dot com.[/b]

[b]When submitting solutions please have the subject in the form Level [X]: Round [X]: [Name]: [Grade][/b]

One point will be deducted for solultions not in that form.

Check out the JMC site for more information:
http://yashar.biz/jmc/

Released on 4/1/05 @ 12:00 AM Eastern Time. Solutions Due by 4/8/05 @ 12:00 AM Eastern Time.

Sending solutions to round one by April 3rd will get you a bonus point!

Number 1 is the same questions as [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=32052]here[/url], so you might want to change it.

Didn't notice...It seems like great minds think alike.  Oh well...

I don't really mind.  Since non intelligent resources are allowed its alright. :D  :D

Sending in answers before tomorrow gets another point!  :D

On your site it says to send them in by April 3. Does that imply that we can send them in tomorrow? (and still get the point.)

Im thinking about giving another day to Level One...

I picked April 3rd because it is my birthday!

So I guess I'll be nice and give one more day  :P

Thanks, and

[color=red] [size=150][b]HAPPY BIRTHDAY!!!!!!!!![/b][/size][/color]

Erg, I just figured about this test about 15-30 minutes ago and submitted my answers for level 1 :? ...

I wish it would be stickied...I'm thinking more people would participate if it was.

What do you mean?  What question is bad?

I think #8 because I wasn't sure if the water was 99 percent of the weight or 99 percent of the watermelon in which case the water could have been anything... Um I might be wrong... Well I should have assumed former in which there actually was a definite answer...

[quote="tarquin"]One of the reasons why I didn't do it was  that I thought one of the questions was bad, and I was like "umm don't feel like doing it".[/quote]
which one?

Then you should have asked! That's why I am here!

I don't really see how much clearer it could get... the watermelon that is 100 pounds is 99 percent water.

So the 99% applies to the weight aswell because 100 lbs=the watermelon  :lol:

Well the other 1 percent could have a different density so... that's why....

I also thought the question about the table and wrestlers was a bad one, but maybe that's just me, and I told you about that in my email of the answers. Did you get it?

When we will know how we did on the test? And about the wrestlers one, it's a bit tricky but not that hard. I don't know if it can be really considered as a "good" problem but it was OK.

[quote="chess64"]I don't think i understood that wrestler one... :?[/quote]

I found it pretty easy....and i thought it was a good problem too

[quote="Farazy"]It was a mock of a problem from a previos mathcounts problem.  I altered it a little though...

I recieved your JMC element  :D[/quote]Thanks!

And the wrestler one is a good problem, but could have been stated more clearly... Either that or I think it needs more clarification because I scrutinize/analyze everything.  :)  :)  :lol:

JMC Round One finished yesterday

Here are the results:
http://yashar.biz/jmc/jmc_files/Level%20One%20Scores.pdf

And here are the solutions:
http://yashar.biz/jmc/jmc_files/JMC%20Level%20One%20Round%20One%20Solutions.pdf

 Good job to all that competed.  :D

Round two is coming soon!  :P

Wait a second... I entered... did you not get my answers? I mailed them to you on like the 4th...

Sorry, I mean the 6th... but I sent them in!

I forgot to turn them in..... :( , but i believe i got 8 correct

Tarquin you scored a 26...Didnt like your last three solutions or the one for number 3

Eh what's wrong with completing the square? :D

Will somebody who can access the PDF file copy and paste it on this page, please?

I have dial up connection, and it is too slow to load the page with the solutions.

Thanks!!!! :)  :)  :)

Farazy, do you know when the JMC level one round two questions will be posted? Thanks.

And with the topic being brought back up, never mind. I was able to get the document to load.  :)

[quote="Farazy"]

Round two is coming soon!  :P[/quote]

How soon?  :D


number theory
greatest common divisor
prime factorization

[quote="chess64"]Suppose that $x, y, z$ are positive integers such that $\frac{1}{x} - \frac{1}{y} = \frac{1}{z}$. Let $g=\text{gcd}(x, y, z)$. Prove that $gxyz$ and $g(y-x)$ are both perfect squares.[/quote]

[hide]
well.... i couldnt prove the first one, so i will assume that it has been ;)

Then $gxyz = k^2$
$\frac{gxy}{z} = \left(\frac{k}{z}\right)^2$ ------------- 1.

From the given condition,
$\frac{y-x}{xy} = \frac{1}{z}$
$z(y-x)=xy$
$\frac{xy}{z} = y-x$

substituting that into 1., we get the desired result, that $g(y-x) = \left(\frac{k}{z}\right)^2$

so someone prove the first one. without it my proof is invalid :D [/hide]

[hide="And here is the first part"]Suppose $g=1$. Then $x,y,z$ cannot share any prime factors. Suppose, by way of contradiction, $xyz$ is not a perfect square. Then in its prime factorization, there exists a prime $p$ such that the highest power of $p$ that divides $xyz$ is odd. Let this highest power be $2n+1$. Therefore, one of $x,y,z$ is not divisible by $p$, another has maximal power $p^e$ (where $e$ can also be zero), and the final one has maximal power $p^{2m+1-e}$. Let $m=\max(e, 2n+1-e)$. Note that the condition implies \[ yz-xz=xy.  \] Hence, if $p^m$ divides one of the variables, it must divide the product of the other two variables. This is quite clearly a contradiction, since $m$ is maximal and $2n+1$ is odd. The conclusion is that $xyz$ is a perfect square.

Suppose $g\neq1$. Then take the substitutions \begin{eqnarray*}a&: =&\frac{x}{g}\\b&: =&\frac{y}{g}\\c&: =&\frac{z}{g}\end{eqnarray*} so that $a,b,c$ satisfy \[ \frac{1}{a}-\frac{1}{b}=\frac{1}{c},  \] and $\gcd(a,b,c)=1$. Hence, $abc$ is a perfect square, from the paragraph above. Therefore, \[ abc=\frac{xyz}{g^3}  \] is a perfect square. Then of course \[ \left(\frac{xyz}{g^3}\right)(g^4)=gxyz  \] is a perfect squre, and the proof is complete?[/hide]

we can prove that the general solution are in fact

$(x,y,z)=(ab,a(a+b),b(a+b))$ for $a,b\in\mathbb{N}$

 ;)


inequalities
function
inequalities unsolved

[b][color=#0000FF]Pro 1[/color][/b]Let $ a,b,c$ such that
$ \sqrt {3}\leq c\leq min(a,b)$
and $ a \plus{} b\geq 5, b\geq\sqrt {5}$
Prove  $ \frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2}\leq\frac {1}{(5 \minus{} \sqrt {5})^2} \plus{} \frac {1}{5} \plus{} \frac {1}{3}$

[b][color=#0000FF]Pro 2[/color][/b] . Let $ x,y>0, x \plus{} y \equal{} 1$
Prove   $ (\sqrt {x} \plus{} \sqrt {y})(\frac {1}{\sqrt {x \plus{} 1}} \plus{} \frac {1}{\sqrt {y \plus{} 1}})\geq\frac {4}{\sqrt {3}}$
 :lol:  :P

[quote="mufc"] Let $ x,y\geq 0, x \plus{} y \equal{} 1$
Prove   $ (\sqrt {x} \plus{} \sqrt {y})(\frac {1}{\sqrt {x \plus{} 1}} \plus{} \frac {1}{\sqrt {y \plus{} 1}})\geq\frac {4}{\sqrt {3}}$
 [/quote]
Try $ x\equal{}1$ and $ y\equal{}0.$ :wink:

But the following inequality:
\[ \left(\sqrt x \plus{} \sqrt y\right)\left(\frac {1}{\sqrt {x \plus{} 1}} \plus{} \frac {1}{\sqrt {y \plus{} 1}}\right)\leq\frac {4}{\sqrt3}
\]
is true with same conditions. :wink:
See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=170430
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=170432

OK. Here is my proof:
since, $ f(x) \equal{} \sqrt x$ is concave function then
$ \frac {1}{\sqrt {x \plus{} 1}} \plus{} \frac {1}{\sqrt {y \plus{} 1}}\leq2\sqrt {\frac {\frac {1}{x \plus{} 1} \plus{} \frac {1}{y \plus{} 1}}{2}} \equal{} \sqrt {\frac {6}{(x \plus{} 1)(y \plus{} 1)}} \equal{} \sqrt {\frac {6(x \plus{} y)}{(2x \plus{} y)(2y \plus{} x)}}.$
Id est, it remains to prove that $ \left(\sqrt x \plus{} \sqrt y\right)^2\cdot\frac {6(x \plus{} y)}{(2x \plus{} y)(2y \plus{} x)}\leq\frac {16}{3}.$
If $ x \equal{} 0$ or $ y \equal{} 0$ then our inequality is obviously true.
Let $ x\neq0,$ $ y\neq0$ and $ \sqrt {\frac {x}{y}} \plus{} \sqrt {\frac {y}{x}} \equal{} u.$
Hence, $ \left(\sqrt x \plus{} \sqrt y\right)^2\cdot\frac {6(x \plus{} y)}{(2x \plus{} y)(2y \plus{} x)}\leq\frac {16}{3}\Leftrightarrow\frac {(u \plus{} 2)u}{2(u^2 \minus{} 2) \plus{} 5}\leq\frac {8}{9}\Leftrightarrow$
$ \Leftrightarrow7u^2 \minus{} 18u \plus{} 8\geq0\Leftrightarrow(u \minus{} 2)(7u \minus{} 4)\geq0,$ which true because $ u\geq2.$


can someone please tell me how to make truth tables in LaTeX?
thank you

Turth tables are just ordinary tables with the appropriate symbols in them. The easiest way in Windows to get the LaTeX code for tables is to use [url=http://www.g32.org/latable/]LaTable[/url]. Use the array option since you are likely to want symbols in the tables.

And here's a quick example:
$ \begin{array}{c|c|c}
 p & q & p\lor q\\
\hline
 T & T & T \\
 T & F & T \\
 F & T & T \\
 F & F & F
\end{array}$
You can use tabular instead of array but then you need to use dollar signs for symbols.
[code]\begin{array}{c|c|c}
 p & q & p\lor q\\
\hline
 T & T & T \\
 T & F & T \\
 F & T & T \\
 F & F & F
\end{array}[/code]

one more question: how do you get the symbol for "not proper subset."  I know that proper subset is $ \subset$ but cant get the line through it.

A line through \subset would be $ \not\subset$ \not\subset but that means [i]not a subset[/i]. What you want is [i]subset but not the whole set[/i] which is $ \subsetneq$ \subsetneq or $ \subsetneqq$ \subsetneqq.
\subsetneq shows up better in a document than it does on the forum as it gets cut off at the bottom. In a document it looks more like the one in $ \subsetneq \frac {1}{2}$ (which is rubbish I know but it forces the whole of the image to show).

the $ \not\subset$ does not seem to be working in display math

umm... never mind i got it THANKS A LOT

You need the amsmath package which you should always have along with amssymb.

ok i got it thanks very much for your help!!!

how do you write the P in power sets like what font is it in

There are a number of possibilities:
1. $ \mathcal{P}$ \mathcal{P}
2. $ \wp$ \wp
or
3. \usepackage{mathrsfs} then \mathscr{P} which gives  [img]http://sixthform.info/gblog/pictures/6d97f9f3e4f6bccb4c48897434b9cb4c.gif[/img]
4. \usepackage[mathscr]{euscript} then \mathscr{P} which gives [img]http://sixthform.info/gblog/pictures/powerset.gif[/img]

what font is in the intermediate counting and probability book

I suggest you ask in the thread about the book http://www.artofproblemsolving.com/Forum/viewtopic.php?t=168015 but you'll get some idea by looking at the pdfs listed at Excerpts at http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=302#excerpt and then looking at File, Document Properties, Fonts.


My question is easy, how did fermat decide the extremum for the equation y = 3 - 2x - x^2

Thanks!

Possibly by writing $y=4-(x^2+2x+1)=4-(x+1)^2$ and then relying on the trivial inequality?


Some references regarding  [b] P. Levy  universal chord theorem.[/b]

In connection with  "Rolle Theorem" : there is a result called [i] "Universal Chord Theorem"[/i] or  [i] "Rolle Theorem for continuous functions".[/i] This is:

[b] Theorem ( Paul Levy see [4],1934) [/b].
[i]If $f \in C[a,b]$  with $f(a)=f(b)$, then for every $h$ of the form $h=\frac{b-a}{n},n \in \{1,2,...\},$ the equation $f(x+h)-f(x)=0\; , \; x \in I: = [a, b-h] ,$ has at least a solution. [/i]

[b]  Proof. [/b]Suppose that $g(x): =f(x+h)-f(x)=0$ has no solution in $I$ . Because $g\in C(I)$  we have  $g(x) > 0$  or $g(x) < 0$ on $I.$ Assume $g(x) > 0\hsp ,\hsp \forall x \in I.$ Because the points $x_{k}: = a+\frac{k(b-a)}{n}\; ,\; \; k \in \{0,1,...,n-1\}, \;$ are in $I$ we must have $\; g(x_{k})= f\left(x_{k+1}\right)-f\left(x_{k}\right) > 0 \; \; \mbox{for}\; \; k \in \{0,1,...,n-1\}.$ By summing  we obtain $f(x_{n})-f(x_{0}) = f(b)-f(a) > 0$ which is a contradiction.

[b]Paul Halmos (see for instance [1]) has proved: let $n\in \{1,2,....\}$ and  $h \ne \frac{b-a}{n}.$ Then there exist $g\in C[a,b]$ with $g(a)=g(b)$ such that equation $g(x+h)-f(x)=0\; , \; \; x \in I: = [a, b-h] ,$ has no solution.[/b]

[b][u] References:[/u][/b]

[1] R. P. Boas , [i] A primer of real functions ,[/i]           The Carus  Mathematical Monographs,            MAA,  vol.13, 1960.

[2] H. Hopf , [i]Ueber die Sehnen ebener Kontinuen und die                Schliefen geschlossener Wege ,[/i]                Comment.Math. Helvetici 9 (1937), 303-319.

[3] R. J. Levit , [i] The finite difference extension of                  Rolle` s  theorem,[/i] Amer. Math. Monthly 70 (1963) , 26-30 .

[4][b] P. L\'evy, [i] Sur une g\'en\'eralisation du theoreme de Rolle,[/i]              C.R. Acad. Sci. Paris 198 (1934) , 424-425.[/b]

The only pronunciation I've heard of "Rolle" is "roll." Also, I believe this appeared in USA TST 2002. I don't know why TST has so many classical results.

The problem that appeared in USA TST 2002 was different.  It was about piecewise linear paths from (0,0) to (n, 0), rather than graphs of continuous functions (in fact, I believe the statement is true of any continuous curve).  I don't know any proof of the Universal Chord theorem that carries over directly.  However, I'd be interested to hear if people have sources, interesting proofs, etc, for the more general problem.

flip2004, I think you meant f(x+h)-f(x), not g(x+h)-f(x).


Finitely many cards are placed in two stacks, with more cards in the left stack than the right. Each card has one or more distinct names written on it, although different cards may share some names. For each name,we define a shuffle by moving every card that has that name written on it to the opposite stack. Prove that it is always possible to end up with more cards in the right stack by picking several distinct names, and doing in turn the shuffle corresponding to each name.

I found [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=231924]this thread[/url] by searching for "USSR shuffle" (without quotes), and searching for "USSR 1968" would also have found it.  That thread has a link to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=225316]this thread[/url], which has a solution.


If you send the solutions, you'll get your score within a day or so.

So far scores: (Only qualified ones so 7 or above)

Xantos C. Guin 10
DarkKnight 8

Can one of the moderators make this topic and the COPSD questions topic stickys? Thanks.

er...Isn't the current one the 2nd one...so why is this so old?

These are the qualifiers from COSPD 2ndA not B.

wait... i thought COP 1st round had to be all correct in order to qualify?


geometry
probability
AMC
USA(J)MO
USAMO
AIME
number theory
\/closed

I'm going to be in precalc honors in 9th grade and I'm wondering if WOOT will contain far too advanced subject material (my geometry class was very basic- nowhere near the "sample woot problem")

So far I've done: Intro to number theory, counting and probability, and geometry

I'm starting on vol.2, which I plan to finish this summer.

I think my AMC 10/AIME scores are currently in the 120/4-7 range, and I'm also concerned about whether the difficulty of the course would make it too time-consuming.

About how much time does WOOT take up, how difficult is it, and is it only helpful towards the olympiad competitions?

I could really use these answers before July 15 (:

The amount of time you put into WOOT is entirely up to you.  That being said, you might get more out of it next year.  If you aren't safely qualifying for USAMO, then the course risks being a waste of time and money.

LawOfSigns, think you're currently right on the fence.  Whether or not WOOT is right for you will probably depend on how much of the AoPS Vol 2 you get through before WOOT starts (and how thoroughly you work through the parts you complete).

You can drop WOOT if you do so by October 5, and get a full refund.  So, what you might want to do is to sign up, then get to work on Volume 2.  If you get to the first lectures of WOOT and find you haven't had much time to work on Volume 2 and the first two classes are way over your head, then drop and take it next year.

WOOT will cover both AIME and Olympiad topics, so it should help qualify for USAMO.

As for time, as t0rajr0u notes, that's up to you.  You should aim to spend 4-5 hours a week on it (including classes), but could spend more if you wanted to.  It's also largely structured so that you can spend more time when you have more time, and less when you don't.  For example, you have several weeks to do the problem sets, so if you have a lot of time one week and no time the other weeks, you can do most of your work on your free week.

Timewise, I actually found that I was regularly attending the classes and was doing all of the problems in the fall.  Then around the end of the fall trimester, work piled up immensely, and I missed a class entirely.  Then I found that, during the winter, I couldn't go for all of the classes because I had track.  Nevertheless, I read the transcripts and did the problems on my own time, and the practice tests/problem sets helped me out.  I also have to say, having an active WOOT forum was much more helpful than I had anticipated: you can make up classes you've missed and learn things very quickly.

Would a 7 or 8 on practice AIMEs be good enough to consider taking WOOT? The AIMEs I took weren't Mock, but past years, and the average is around 7 or 8.

I'd probably be extremely busy next year though, so 4.5 hours a week might be a bit hard. Would it still be worth it if I want to participate in USAMO next year?

Well, I seem to recall Mr. Rusczyk saying that WOOT is for people who can confidently score at least a 5 on the AIME, so a 7 or 8 seems to be good.

In my experience, scores on practice tests can vary greatly from scores on actual tests. Also, what you get out depends on what you put in. If you sign up and never go to a class or do the homework, then you won't get anything out. In that sense, whether  or not it will be worth it is for you to decide.


[size=150]Mandelbrot march 1992[/size]

4. Bill and Ted are playing a most excellent game. They begin with $39$ sticks; each player on his turn removes at least one but no more than five sticks. The loser is the one forced t take the last stick. If Bill goes first, then how many sticks should he take to be sure of winning?

[hide]If Bill takes two sticks, there will be 37 left. Then whenever Ted takes $s$ sticks, Bill can just respond by taking $6-s$ sticks. Then each time both go, 6 will be removed. After this happens 6 times, 1 stick will be left and it will be Ted's turn to take the last stick!

Therefore Bill can win by taking $\boxed{2}$ sticks on the first turn.[/hide]

[hide="Beautiful problem"]Well, after Bill's last tum he wants there to only be $1$ stick. Now for each meow number $x$ of sticks Ted takes, Bill can take $6-x$ sticks. Bill should go to one stick, so he'd need to get to $1+6=7,7+6=13, 19, 25, 31,37$ sticks left somehow. To get to $1,7,13,19,25,31$ sticks left, he could use my strategy, to get to $37$ sticks, he needs to take $\boxed{2}$ sticks in the beginning.[/hide]


[hide]rewrite it as 5(3/5 sin x + 4/5 cos x). Let a=arcsin(4/5). Then the expression is 5 sin(x+a).  The amplitude, and the maximum, is 5.[/hide]

That is correct. ( That's a nice method. I didn't think of that one...)

A different method:

[hide]By Cauchy-Schwarz,

(3sinA+4cosA)^2 :le: (3^2+4^2)(sin^2A + cos^2A)

(3sinA+4cosA)^2 :le:25.

Thus the maximum is sqrt[25], or 5. [/hide]

[hide]

Wow, I'm stupid. I thought it said 3sinA+4sinA. wow.

[/hide]


calculus
integration
function
real analysis
real analysis unsolved

Suppose $ f(x), f_n(x)$ are integral functions on a measurable set $ E$, $ f_n \to f$ a.e. on $ E$ and $ \int_E |f_n| \to \int_E |f|$ as $ n \to \infty$. Show that for every measurable subset $ A$, $ \int_A |f_n| \to \int_A |f|$ as $ n \to \infty$.

Any hints?


The sum of three different prime numbers is 40. Find the product of the largest two of the primes.

Clearly, $ 2$ must be one of these primes, or the sum would be odd.

Therefore, $ p_1\plus{}p_2\equal{}38$.

$ 31\plus{}7\plus{}2\equal{}40$, so $ 31 \cdot 7\equal{}217$.


linear algebra
matrix
IMO Shortlist
algebra
polynomial
system of equations
IMO Longlist

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

[quote="orl"]Solve the system of equations:

$ \begin{matrix} x^2 + x - 1 = y \\ y^2 + y - 1 = z \\ z^2 + z - 1 = x. \end{matrix} $[/quote]

Let (x, y, z) be a solution of this system. Then, let a = x + 1, b = y + 1 and c = z + 1. Then, x = a - 1 and y = b - 1, so that the equation $ x^2+x-1=y $ becomes $ \left(a-1\right)^2+\left(a-1\right)-1=b-1 $, or, equivalently, $ a^2-a-1=b-1 $. Thus, $ a^2-a=b $, so that $ a^2=a+b $. Similarly, $ b^2=b+c $ and $ c^2=c+a $. Hence, $ \left(a-b\right)a^2=\left(a-b\right)\left(a+b\right)=a^2-b^2=\left(a+b\right)-\left(b+c\right)=a-c=-\left(c-a\right) $. Similarly, $ \left(b-c\right)b^2=-\left(a-b\right) $ and $ \left(c-a\right)c^2=-\left(b-c\right) $. Multiplying these three equations, we get

$ \left(a-b\right)a^2\cdot\left(b-c\right)b^2\cdot\left(c-a\right)c^2=\left(-\left(c-a\right)\right)\cdot\left(-\left(a-b\right)\right)\cdot\left(-\left(b-c\right)\right) $.

Now, if the numbers b - c, c - a and a - b are all nonzero, then we can divide this equation by (b - c) (c - a) (a - b) and obtain $ a^2b^2c^2=-1 $, what is impossible, since $ a^2b^2c^2\geq 0 $ (as squares are always nonnegative). Thus, the numbers b - c, c - a and a - b cannot all be nonzero; one of them must be zero. WLOG, assume that b - c = 0. Then, b = c, so that y = z (since b = y + 1 and c = z + 1). Hence, the equation $ y^2+y-1=z $ becomes $ y^2+y-1=y $, so that $ y^2=1 $. Thus, either y = 1 or y = -1.

If y = 1, then y = z yields z = 1, and thus $ x=z^2+z-1=1^2+1-1=1 $, so we get the solution (x, y, z) = (1, 1, 1). A trivial check shows that the triple (x, y, z) = (1, 1, 1) is indeed a solution of our system of equations.

If y = -1, then y = z yields z = -1, and thus $ x=z^2+z-1=\left(-1\right)^2+\left(-1\right)-1=-1 $, so we get the solution (x, y, z) = (-1, -1, -1). A trivial check shows that the triple (x, y, z) = (-1, -1, -1) is indeed a solution of our system of equations.

Hence, our system of equations has the two solutions (x, y, z) = (1, 1, 1) and (x, y, z) = (-1, -1, -1).

  darij

I can solve this problem for all $x,y,z\ge0$
suppose that $x\ge y \Rightarrow x^2+x-1\ge y^2+y+1$ so we have $y\ge z$ and again we have $z\ge x$ hence
$x\ge y\ge z\ge x \Rightarrow x=y=z$

and if we had $x\le y \Rightarrow x^2+x+1\le y^2+y+1$ so $y\le z$ and now we have $z\le x$ and again
$x\le y\le z\le x \Rightarrow x=y=z$
so if we have $x,y,z\ge 0 \Rightarrow x=y=z$ and the only thing we haveto do is tosolve this equation
$x^2+x-1=x \Rightarrow x^2=1 \Rightarrow x=y=z=1$

but if $x,y,z\not\ge 0$  :blush:
I tried to use a similar solution of this system of equations
$(x+y)^3=z$
                                                                                $(y+z)^3=x$
                                                                                $(z+x)^3=y$
but it didn't work for $x,y,z\not\ge 0$   :?:  would you please help me to complete this solution.
and the solution of darij was really beautiful.
thanks...

I got it through a slightly different method but yeah darij  your solution really is beautiful
 Ashwath

Orl when all problems will be solved, could you post the solutions at PDF like the problems? Thanks

[quote="pkothari13"][quote="darij grinberg"]
Let (x, y, z) be a solution of this system. Then, let a = x + 1, b = y + 1 and c = z + 1. Then, x = a - 1 and y = b - 1[/quote]

Nice solution! But one question - how did you think to come up with those substitutions?[/quote]

Yeah how?Any trick or just practice?

[quote="pkothari13"][quote="darij grinberg"]
Let (x, y, z) be a solution of this system. Then, let a = x + 1, b = y + 1 and c = z + 1. Then, x = a - 1 and y = b - 1[/quote]

Nice solution! But one question - how did you think to come up with those substitutions?[/quote]

I just looked for a substitution of the kind a = x + k, b = y + k, c = z + k which would make the equations
$\begin{matrix} x^2 + x - 1 = y \\ y^2 + y - 1 = z \\ z^2 + z - 1 = x\end{matrix}$
not contain the nasty 1's, but just contain the variables a, b, c. And found that the right k is 1.

  Darij

I also used the substitution $a=x+1$, but not in the same way.

Let $a=x+1, b=y+1, c=z+1$. We then have the system
$\begin{matrix}xa=b \\ yb=c\\ zc=a\\ \end{matrix}$
which gives us (assuming $a,b,c\not =0$, I'll come back to that one)
$xyz=1$

In a trivial way (add the three eequations to each other), we also find that the system
$\begin{matrix} x^2 + x - 1 = y \\ y^2 + y - 1 = z \\ z^2 + z - 1 = x. \end{matrix}$
gives us $x^2+y^2+z^2=3$.

Since $xyz=\frac{x^2+y^2+z^2}{3}=1$, we have by the AMGM-inequality that $x=y=z=1$ and (if we include negative numbers, which we do) $x=y=-1, z=1$ and permutaions of that. But since we here assumed that $a,b,c\not=0$ which is equivalent to $x,y,x=-1$, we can throw away that solution (it is false).

Now assume one of $a,b,c$ equals $0$. We then have (assuming because of symmetry that $a=0$, or $x=-1$)
$\begin{matrix} - 1 = y \\ y^2 + y - 1 = z \\ z^2 + z = 0. \end{matrix}$
and it is obvious that the only solution to this equation is that $x=y=z=-1$

Therefore the two solutions are $x=y=z=1$ and $x=y=z=-1$

Hopefully I haven't made too many errors... Edit: Fixed one minor I found.

I only noticed that (-1;-1;-1) works. My solution is based upon the cartesian representation and the growing and the functional equation f^3(x)=x with f(x)=x^2+x-1 and the roots of the polinomial and derivatives...

Add the equations to get

$x^2+y^2+z^2=3$

Rewrite the equations as

$x(x+1)=y+1$
$y(y+1)=z+1$
$z(z+1)=x+1$

If $x,y,z\neq -1$, multiplying the equations we get

$xyz=1 \Rightarrow x^2y^2z^2=1$

You can easily check none of $x, y, z$ is 0, thus $x^2,y^2,z^2 \in R^+$

And by AM-MG

$1=\frac{3}{3}=\frac{x^2+y^2+z^2}{3}\geq \sqrt[3]{x^2y^2z^2}=\sqrt[3]{1}=1$

We have the equality, thus $x=y=z=1$

If some of $x, y, z$ is $-1$, it´s easy to see $x=y=z=-1$

Thus, the solutions are $x=y=z=t \in$ {$1, -1$}


Let n be a positive integer.
Prove that the following two statements are equalient:
-n is not divisible by 4.
-There exist integers a,b such that a^2+b^2+1 is divisible by n.


inequalities
function
integration
real analysis
real analysis unsolved

Hi,

I have a equation as follows:

Let $ A$ be finitely measurable, and $ 1\le p<q<\infty$. Without applying Holder inequality, prove $ L^q(A)\subset L^p(A)$.

Thanks.

Sketch of the proof :first observe that it's enough to prove this for nonnegative functions. Now if one has the fuction $ f$ with $ \int f^q < \infty$ we can denote $ \alpha \equal{} \frac {p}{q}$ , $ g \equal{} f^q$. Let's define  sets $ A_n \equal{} (x\in X| n\le g(x) < n \plus{} 1)$. We can now estimate: $ M \equal{} \int g \le \sum n\cdot m(A_n)$. On the other hand $ \int f^p \equal{} \int g^{\alpha} \le \sum (n \plus{} 1)^{\alpha} m(A_n) \le m(X) \plus{} M$ because $ \alpha < 1$.

Or simpler. (As leshik did, I'm assuming that $ f\ge 0$.)

Note that if $ f>1$ at some point, then $ f^p<f^q$ at that point.

$ \int_Af^p\equal{}\int_{\{f>1\}}f^p\plus{}\int_{\{f\le 1\}}f^p$

$ \le\int_{\{f>1\}}f^q\plus{}\int_{\{f\le 1\}}1$

$ \le\int_Af^q\plus{}\int_A1\equal{}\|f\|_q^q\plus{}m(A)<\infty.$

Of course, this estimate isn't particularly quantitative or "tight", and doesn't scale well. It only shows the required set inclusion. If you want tight estimates, then you use Hölder.

Hi leshik and Kent Merryfield,

It is terrific. Thank you very much.


I noticed that on the All Time Top 10 list there is a category called "Alcumus Score."  What is it?  How is it calculated?

It's calculated by our super-whizzy Alcumus algorithm (it really is super-whizzy -- backed by a lot of fancy AI stuff with intimidating sounding names).  It gives each person and each problem a score based on how each person has done on each problem that person has tried to solve.

Hmm...how can we see our Alcumus score?

Also, my percentage is higher than one of my friend's, but his/her Alcumus score ranking is 14 higher than mine...what's wrong?

The Alcumus score takes into account the difficulty of the problems you have seen.  Your friend has likely been confronted with harder problems than you have.

Next to "Alcumus Score:" under the report page, it says "What's this?".

not that there is a grey box that appears when you hover your mouse over the "what's this"
at the bottom, it says "Note that the Alcumus Score is not updated in real time"

i'm guessing you can't see it because of this, but i'm not sure...

No one answered to Dojo?
It's a good question, when and how can we view our Alcumus Score?
Only thing I see is my position when everyone is sorted by Alcumus score.

This will be displayed on your report page during our next code commit. There was a bug preventing it from being displayed for many users.

could one of you please tell me the formula for the Alcumus score?

I think it is probably a secret formula.

Edit: Thank you, claudiafeng, for PMing me that I spammed here, and so now, I have edited it.

I have just known today that the user who goes by the name of CutePenguin has brought to life an old topic, while creating this post.

Also,[size=50]I predict that this post has low rating.[/size]

I assume that dantx5 thinks that his post will have a low rating. I brought this old topic to life because I think that this is important, as a lot of people are curious about this, and rruzyck was able to answer this at the start, so people will stop making new topics about this.

Why base it on how well other people have done? It's like a knowledge game to see how much you know.


Functional Analysis
real analysis
real analysis unsolved

I don't know if it is classic but it looks nice.

Let $(A_{n})_{n \in \mathbb{N}}$ be decreasing sequence of nonempty, closed, convex and bounded sets in a Hilbert space. Prove that $\bigcap A_{n}\neq \emptyset$.

It is false in Banach spaces (e.g. easy countexamples in $c_{0}, \ell_{1}$).

Any ideas how to prove it?

Prove that if the intersection is empty, than the distance from any point $x$ to $A_{n}$ tends to infinity. In order to do this, it is useful to note that $|p_{U}(x)-p_{V}(x)|^{2}\leq d(x,U)^{2}-d(x,V)^{2}$ if $U$ is a subset of $V$ and both are closed convex sets.

More is true:

If $\mathcal A=(A_\alpha)_\alpha$ is a collection of non-empty closed, convex, and bounded subsets of a Hilbert space which has the finite intersection property (that is, every finite subcollection has non-empty intersection), then $\mathcal A$ has non-empty intersection. This follows immediately from the fact that $A_\alpha$ are weakly compact.

P.S.

Clearly, the argument above works for any reflexive Banach space.

I have constructed a sequence as follow:
take x in A1 and project it in A2
we can suppose that  the projection doesnt belong to A3
So project in A3 and , construct the sequence
I have shown that if the limit exist of this sequence then it belongs to the full intersections
But i failed to prove the convergence of the sequence.Does this sequence exist.


$f: R \implies R$ and $f$ is bounded below, continuous function. Prove that for all $x \in R$ and  $x \ne x_{0}$,  $\exists x_{0}$ so that $f(x_{0})-f(x)<|x-x_{0}|$ .

Take $f(x)=|2x|$. For $x=0$, does there exist any $x_{0}\neq 0$ such that $f(x_{0}) = 2|x_{0}| < |x_{0}|$ ?
I think there's a mistake in the wording of your problem.

to julien-satini,in your ex. take X0=0,so you cant take X=0.
 idea is to choose X0 as y for which f(y)+mody is minimum. :D

To who. Do you know solution? If yes, can you post it?

@ who: maybe, but we must have $x \neq x_{0}$ according to the original post, so I cannot set $x_{0}=0$, can I?

julien we have to fix x0 not x.
to mayhem,i  have sol. take x0 as mentioned in prev. post. use tri, ineq.
to show such minimaexists isnt hard.than show such points of minima form the interval if minima isnt unique.f will have mini. on that ointerval take such point of minima,it workkks! :D


calculus
derivative
calculus computations

Well I'm in pre-calc but our teacher gave us a calculus problem that we shouldn't know how to do and is offering bonus points to anyone who can figure it out. I'll just type it up verbatum.

One dark rainy night Kim and Adam arrived at their job as air traffic controllers at Dayton international airport. There were only two airplanes flying in the direction of Dayton at the time. One airplane was a 747 jet approaching from 400 miles to the east at 325 MPH. Its altitude was 10000 ft and descending at 70ft/min. The other plane was a 10 passenger prop plane approaching from a point 260 miles to the south at 211.25MPH. The second plane's altiture was 3000 feet and climbing at 24.8 ft/min.
Suddenly the computers all lose power. Kim and Adam are equipped with only their TI's. An accident is considered likely if the two aircraft pass within 200 ft of eachother.
Find the minimum distance between the airplanes are when this will occur.

So right now I have
f(x)=sqrt(((10000-4200x)-(3000+1488x))^2+(400-325x)^2+(260-211.25x)^2)

the test is x=3 y=685.797124204

I don't want somone to flat out tell me the answer but if you could help tell me what I'm doing wrong.

[quote="tetrahedr0n"]shib, wouldn't it be unfair to the other students in the class if we told you what you were doing wrong and you solved the problem? Will your teacher still accept your problem if he or she knew you had help from others?[/quote]

If I was the teacher (or other  student) I will NOT consider it unfair  (unless, of course  it was mentioned  not to  get outside help). In fact  i will be happy if my student got to learn new things. (In fact I may, if asked,  will myself give hint or help the student). After all the whole aim is to teach.  I would actually encourage students to discuss the problem among themselves of get help if that  aids to learning .

Okay back to problem: Here are some hints:

[hide]1. Instead of distace, take square of the distance.. so that you get rid of the messy square-root sign. After all if the square is minimum so would its square-root.
2.  If you know calculus,  use standad way to get the minimum (eg equate the derivative to  zero etc..).
3.  If you don't know calculus -  trying to complete the square  (and remember square of any number is always positive) will do the trick ... ( minimum value will be reached when "that"  is equal to zero etc ...)  The hint is almost give away for the whole problem.

Good luck ..[/hide]

Go as far as you can without substituting numbers. Before you substitute numbers, you have to convert all quantities to the same units (such as miles everywhere for distances, hours everywhere for time, miles/hour for velocities). Otherwise, you used the correct equation. If you minimize using derivatives, notice that the distance is minimum iff the square of the distance is minimum. Minimize the square for easier differentiation to find the time for the minimum distance.

[color=red][b]Plane 1[/b][/color] (Boeing 747):

$x_{01} = +400\ mi$

$v_1 = -375\ mi/hr$

$z_{01} = +10000\ ft = \frac{10000}{3 \cdot 1760}\ mi = +\frac{1250}{66}\ mi$

$v_{z1} = -70\ ft/min = -\frac{70 \cdot 3600}{3 \cdot 1760}\ mi/hr = -\frac{525}{11}\ mi/hr$

[color=red][b]Plane 2[/b][/color] (prop plane):

$y_{02} = -260\ mi$

$v_2 = +211.25\ mi/hr$

$z_{01} = +7000 ft\ = \frac{7000}{3 \cdot 1760}\ mi = +\frac{875}{66}\ mi$

$v_{z1} = +24.8\ ft/min = \frac{24.8 \cdot 3600}{3 \cdot 1760}\ mi/hr= +\frac{186}{11}\ mi/hr$

$D^2 = (x_{01} + v_1 t)^2 + (y_{02} + v_2 t)^2 + [z_{01} - z_{02} + (v_{z1} - v_{z2})\ t]^2$

$\frac{d(D^2)}{dt} = 2(x_{01} + v_1 t)v_1 + 2(y_{02} + v_2 t)v_2 + [z_{01} - z_{02} + (v_{z1} - v_{z2})\ t](v_{z1} - v_{z2}) = 0$

$t = \frac{x_{01} v_1 + y_{02} v_2 + (z_{01} - z_{02})(v_{z1} - v_{z2})}{v_1^2 + v_2^2 + (v_{z1} - v_{z2})^2}$

Preliminary calculation:

$z_{01} - z_{01} = \frac{1250}{66} - \frac{875}{66} = +\frac{125}{22}\ mi$

$v_{z1} - v_{z2} = -\frac{525}{11} - \frac{186}{11} = -\frac{711}{11}\ mi/hr$

Pay attention to the signs of $x_{01}, y_{02}, v_1, v_2, v_{z1} - v_{z2}$ when substituting the values. The time comes out in hours and you have to calculate it to about 0.1 sec, i.e., to about 0.0001 hr. The time at the minimum distance is then $t = 1.2242\ hr$. When you substitute this to the formula for the distance square $D^2$ and take the square root, the minimum distance comes out in miles. Convert to feet and you are done.


calculus
integration
calculus computations

hello,

I am doing this example from the calculus 3 book :
Find the mass of the triangular lamina with vertices (0,0), (0,3).and (2,3) given that the density at (x,y) is p(x,y)=2x+y, and I can't understand why they are saying x=2y/3  and they have two integral; first one from 0 to3 and second one from 0 to 2y/3

Page 1009 right?

They chose to integrate with respect to $ y$ in the final integral, so their iterated began with $ dxdy$. You can integrate with respect to $ x$ by beginning with $ dydx$, but you'll have to subtract that from another integral.

$ \int^3_0\int^{\frac{2}{3}y}_0 (2x\plus{}y)dxdy\equal{}\int^2_0\int^3_0 (2x\plus{}y)dydx\minus{}\int^2_0\int^{\frac{3}{2}x}_0 (2x\plus{}y)dydx$

Finding the equation of the line from $ (0,0)$ to $ (2,3)$,
$ \frac{3\minus{}0}{2\minus{}0}\equal{}\frac{3}{2}\Longrightarrow y\equal{}\frac{3}{2}x\Leftrightarrow x\equal{}\frac{2}{3}y$

So I'll do their integral:
$ \int^3_0\int^{\frac{2}{3}y}_0 (2x\plus{}y)dxdy\equal{}A$
We'll just integrate once with respect to x,
$ \int^{\frac{2}{3}y}_0 (2x\plus{}y)dx\equal{}[x^2\plus{}yx|^{\frac{2}{3}y}_0\equal{}\frac{4}{9}y^2\plus{}\frac{2}{3}y^2\equal{}\frac{10}{9}y^2$
Then we'll integrate that with respect to y,
$ \displaystyle \frac{10}{9}\int^3_0 y^2dy\equal{}\left[\frac{10}{27}y^3\right|^3_0\equal{}10$

See? Not a problem. I think you're just confused about the meaning of $ \int_R \int \rho dA$. That R represents a region whose bounds you must define. In this case, it's a triangle with vertices at $ (0,0),(0,3),(2,3)$. Describing that region is where the line $ x\equal{}\frac{2}{3}y$ came from, as well as all of the other upper and lower limits of integration.


Suppose that my car license plate consists of a 3 digit number. if you add 1 to that number , it is divisible by 7. If you add 4 , it is divisible by 8. If you add 7 it is divisible by 9. What is the number?

Hint
[hide]
Use Chinese Remainder Theorem  :)
[/hide]


Let $x,y,z>0$ with $n \in N \neq 0$. Show that:

$ \sum x^{n+1}(y+z) \geq 2xyz( \sum x^{n-1} )$.

cheers! :D :D :cool:

It can be written (sum 1/x)*(sum x^n)>=3 sum x^n-1, which holds by Cebasev.


search
number theory unsolved
number theory

$ a,b,c,d$  $ \in \mathbb{Z^{\plus{}}}$

Prove that many $ (a,b,c,d)$  satisfies this condition;

$ ab\plus{}1,bc\plus{}1,ac\plus{}1,bd\plus{}1,ad\plus{}1,cd\plus{}1$ are perfect square

It posted before  :wink:  [url]http://www.mathlinks.ro/viewtopic.php?search_id=1725516229&t=149466[/url]


Hi.
I was wondering what topics are tested in these questions?qns 1-10 and what books are are a good read to tackle these sort of problems
http://www.maths.usyd.edu.au/u/SUMS/sums2007.pdf

Also, if you want to practice here is a nice site for university students.
http://www.maths.usyd.edu.au/u/SUMS/

solutions are included

Look at Graham White go  :) . He's doing very well even after high school.

Who is he? Personal friend?
He answered alot of the questions

How would this compare to IMO?

Graham White went to James Ruse Agricultural High School and went to the IMO 3 times (Honourable Mention, Bronze, Silver) in 2004, 2005, 2006.

Yep, but I am an accelerated student so I can just do them again if I screw up  :D .

But sticking to the topic, I am not sure whether these problems are Olympiad styled (certainly Olympiad training may help). I think you would need to ask someone more experienced.

The following fields seem to be strongly represented: combinatorics, algebra (both olympiad and undergraduate) and algebraic combinatorics. But taking the 2006 questions in consideration as well, you should also train analysis and number theory. Usually, the problems of such contests are "side products" of mathematicians researching in advanced fields, so learning more wouldn't exactly be a disadvantage, and creative thinking beyound the standard tricks is also of use. It's not a competition you can win by reading Engel's PSS.

  Darij

[quote="darij grinberg"] Usually, the problems of such contests are "side products" of mathematicians researching in advanced fields[/quote]

Can you give an example from this contest or otherwise?

Thanks.

Well, if you try to devise an olympiad problem, you seldomly will come up with things like #7 or #10 - such stuff usually appears as lemmata in advanced algebra (particularly, representation theory produces a lot of combinatorics problems) and advanced combinatorics. Besides, some problems like #4, #5 seem to me directly useful in advanced fields.

  darij


[quote="Wumbate"]This comes from [i]the art and craft of problem solving[/i] (6.2.18):

Find a nice formula for the sum:
(note: nCm means n choose m.. I couldn't find it in latex)
$ nC0\plus{}nC1\plus{}nC2\plus{}...\plus{}nCn$
Bonus: Provide an explanation for why it is true, not just the math.

I've made a table of values but I can't find a nice formula.  I've got an interpretation for what the sum represents, but I don't see any way to 'reinterpret' it to come up with a nice formula.  Help?[/quote]
Latex tip: Use {n\choose m} to produce $ {n\choose m}$

[hide="Easy"]
Use binomial theorem,
$ (1\plus{}1)^{n}\equal{} 2^{n}$
[/hide]

[quote="Wumbate"]
Bonus: Provide an explanation for why it is true, not just the math.
[/quote]
[hide]I think it's $ 2^{n}$ because you can imagine a row of $ n$ light switches and then it is the number of off/on combos for the lights, which can be counted using the combinatorics things or just $ 2^{n}$ because of $ n$ events each with $ 2$ possibilities.[/hide]

[hide="ans"]$ \binom{2n}{n}$[/hide]

[hide="proof"]suppose u had a clollection of $ 2n$  distinct objects and had to choos $ n$ out of them not carrying about the order then the number of ways are $ \binom{2n}{n}$ well u could do this another way
say u divide the $ 2n$ object int two heaps of $ n$ and $ n$ say A and B.
now i can do the choice of $ n$ objects like this
let $ (a,b)$ refer to choice of $ a$ objects from A and $ b$ objects from B.
now i can have my choice as $ (n,0),(n\minus{}1,1),(n\minus{}2,2).....(0,n)$
so the no. of ways have to be same .but in case 2 all events being independent the numebr of ways
are nothing but
$ \sum_{k \equal{} 0}^{n}\binom{n}{k}\binom{n}{n\minus{}k}\equal{}\sum_{k \equal{} 0}^{n}\binom{n}{k}^{2}\equal{}\binom{2n}{n}$
hence we are done
[/hide]

[quote="Wumbate"]I'm sorry, I put the wrong formula down.  Here is the actual problem:

Find a nice formula for the sum:
$ {n\choose 0}^{2}\plus{}{n\choose 1}^{2}\plus{}{n\choose 2}^{2}\plus{}...\plus{}{n\choose n}^{2}$
And the bonus: find a way to interpret the summation two different ways to give an "explanation" for why the formula is correct.[/quote]

[hide="Generating Function"]
Now,
$ (1\plus{}x)^{n}(1\plus{}x)^{n}\equal{} (1\plus{}x)^{2n}$
Write as generating function and let $ x\equal{}1$.
[/hide]


modular arithmetic
number theory proposed
number theory

Let $ k,n$ be natural numbers which are coprime such that $ 1\le k\le n\minus{}1$ and let $ M\equal{} {1,2,3,4,......,n\minus{}1}$. Every element of $ M$ is coloured with one of the two colours such that :
1. $ i$ and $ n\minus{}i$ have the same colour for all $ i\in M$
2. for $ i \in M$ and $ i \neq k$ , $ i, |k\minus{}i|$ have the same colour.
Prove that all emenets of $ M$ are coloured with the same colour.

[hide]If $ k\equal{}1$ it is obvious, so assume $ k>1$. Let $ n\equal{}pk\plus{}q$ where $ p,q\in\mathbb{N}$ and $ 0<q<k$.

$ \{1,k\plus{}1,2k\plus{}1,\ldots,pk\plus{}1,k\minus{}1,2k\minus{}1,\ldots,pk\minus{}1\}$

$ \{2,k\plus{}2,2k\plus{}2,\ldots,pk\plus{}2,k\minus{}2,2k\minus{}2,\ldots,pk\minus{}2\}$

...

$ \{q,k\plus{}q,2k\plus{}q,\ldots,pk\plus{}q,k\minus{}q,2k\minus{}q,\ldots,pk\minus{}q\}$

All elements in each set above have the same colour. $ q\plus{}1$ has the same colour as $ k\plus{}q\plus{}1$ which has the same colour as $ n\minus{}k\minus{}q\minus{}1\equal{}(p\minus{}1)k\minus{}1$ which is an element of the first set. So, similarly, two integers have the same colour if they have the same modulo $ q$.  But $ q\minus{}1$ has the same colour as $ k\plus{}q\minus{}1$ which has the same colour as $ n\minus{}k\minus{}q\plus{}1\equal{}(p\minus{}1)k\plus{}1$ which is an element of the first set. Similarly, $ a$ and $ b$ have the same colour if $ a\equiv \minus{}b\pmod{q}$.

Therefore, all properties of $ k$ mentioned in the problem is the same as $ q$, so we can change $ k$ in the problem into $ q$. Now assume $ n$ leaves a remainder of $ q_1$ upon division by $ q$. Similarly as above, we can all change $ q$ in the problem into $ q_1$. Repeating this procedure, we must reach a point where the value of $ k$ in the problem becomes 1, which is trivial. Hence we are done.[/hide]


geometry
3D geometry
algebra
polynomial
arithmetic sequence
number theory proposed
number theory

An result :
$ f(x)$ is a perfect square for all $ n\in N$ then $ f(x)$ is a square of a polynomial.
Now consider our problem :
$ a_{n_0\plus{}1}\equal{}p^3\equal{}an_0\plus{}b$ then
Imply that $ (p\plus{}ta)^3$ is a term of this sequence. $ t\in N$
Imply that $ n_t\equal{}\frac{(p\plus{}ta)^3\minus{}b}{a}\equal{}f(t)$
Because $ f(t)$ is a polynomial of t with degree 3 .
So it can not is a perfect square for all $ t\in N$
This problem claim.

If $ f(t_0)$ isn't a square then $ (p\plus{}t_0a)^3$ too. Why?

[quote="N.T.TUAN"]If $ f(t_0)$ isn't a square then $ (p \plus{} t_0a)^3$ too. Why?[/quote]
I don't understand ? I have prove that $ f(t)$ contain infinite term not is a perfect square.

[hide="Hint:"]
Let $ n^3$ be a perfect square then $ n$ is a perfect square too.[/hide]

[quote="N.T.TUAN"]An infinite increasing arithmetical progression consists of positive integers and contains a perfect cube. Prove that this progression also contains a term which is a  perfect cube but not a perfect square.[/quote]
arithmetical progression$ (ai \plus{} b)_{ i\in\mathbb{N}}$
we suppose that $ p^3 \equal{} an \plus{} b$ for some $ n\in\mathbb{N}$
then $ \forall t\in\mathbb{N}: \ (p \plus{} at)^3 \equal{} a(n \plus{} 3pat \plus{} 3p^2t \plus{} a^2t^3) \plus{} b\in\{ai \plus{} b: \ i\in\mathbb{N}\}$
if $ a^2a\plus{}p$ and $ a^2(1\plus{}a)\plus{}p$ are perfect squars then $ (a^3\plus{}a\plus{}p)\minus{}(a^3\plus{}p)\ge (a\plus{}1)^2\minus{}a^2$ (impossible)
then $ \exists t\in\{a^2,a^2\plus{}1\}: \ p \plus{} at$ isn't perfect square.
then $ (p \plus{} at)^3\in\{ai \plus{} b: \ i\in\mathbb{N}\}$ a perfect cube but not a perfect square.

I will give a complete solution to this problem:
Let $ A$ be an arithmetical progression and $ a^3\in A$.Consider 2 cases:
1.$ a^3$ is not a perfect square,hence we are done.
2.$ a^3$ is a perfect square then using the lemma(see my post above),we obtain that $ a \equal{} b^2$,for some $ b\in N$.
Also we know that $ (a \plus{} ad^2)^3\in A$,hence if $ (a \plus{} ad^2)$ is a perfect square too,then $ b^2 \plus{} b^2d^2$ is a perfect square too,which is impossible.Contradiction.

Sorry I don't read problem carefully .
I have prove that there exist infinite $ n$ such that $ a_n$ is a perfect cube but n is not a square.
This problem can be general as follow :
[u]Problem 1[/u] Let $ p,q\in N$ where $ q>1$ and $ \gcd(p,q)\equal{}1$
 Prove that if a arithmetic sequence $ a_n$ contain an $ p$ th power then it contain infinite term be a p th power but not a $ q^{th}$ power of integer .
Proof :$ x_{n\plus{}1}\equal{}an\plus{}b$
Let $ k^p$ is a term of this sequence then $ (k\plus{}tb)^p$ also is a term too.
Let chose $ k\plus{}tb$ is not $ q^{th}$ power then it is not a $ q^{th}$ power .
This problem was be claim.


Are there such positive integer numbers $ p$ and $ q$ that p is prime and
$ 3^p\plus{}7^p\equal{}2\cdot5^q$.

Case 1 :$ p \equal{} 2;5$ then easy to check that it doesn't satisfy condition .
Case 2: $ p$ is different  from 5. Then $ q\geq 2$
But we have
$ \gcd(10,\frac {7^p \plus{} 3^p}{10}) \equal{}\gcd(p,10)\equal{} 1$ therefore
$ 25 \not |3^p \plus{} 7^p$
So equation has no solution .

[quote="TTsphn"]
But we have
$ \gcd(10,\frac {7^p \plus{} 3^p}{10}) \equal{} \gcd(p,10) \equal{} 1$[/quote]

Could you please explain it? Unfortunately it's not so obvious to me  :oops:

This is an well know lemma .
If $ a,b>1$ and $ \gcd(a,b)\equal{}1$ then
$ \gcd(a\minus{}b,\frac{a^n\minus{}b^n}{a\minus{}b})\equal{}\gcd(a\minus{}b,n)$

Aaah, thank you. I knew that lemma before, but I didn't see it here. Thank you very much :wink:


Now if $ {x}$ is equal to the sum of two squares show that $ {5x}$ is also a sum of two squares.

Given that $ x \equal{} a^2 \plus{} b^2$, $ a \geq b$.
Then $ 5x \equal{} 5a^2 \plus{} 5b^2 \equal{} (2a \minus{} b)^2 \plus{} (a \plus{} 2b)^2$

This follows from the generalization (note: all variables are nonnegative integers unless otherwise indicated):

[i]If $ u$ and $ v$ are expressible as the sum of two squares, then $ uv$ is also expressible as the sum of two squares.[/i]

[hide="Proof"]
Let $ u \equal{} a^2 \plus{} b^2$ and $ v \equal{} c^2 \plus{} d^2$. Then $ uv \equal{} (a^2 \plus{} b^2)(c^2 \plus{} d^2) \equal{} a^2c^2 \plus{} a^2d^2 \plus{} b^2c^2 \plus{} b^2d^2 \equal{} (ac \minus{} bd)^2 \plus{} (bc \plus{} ad)^2$, and we are done.[/hide]

Since $ 5 \equal{} 1^2 \plus{} 2^2$, the result follows immediately.

[hide="Side note"]
In fact, $ uv$ is expressible as the sum of two squares in [i]two[/i] ways: $ (ac \minus{} bd)^2 \plus{} (bc \plus{} ad)^2$ or $ (ac \plus{} bd)^2 \plus{} (bc \minus{} ad)^2$.[/hide]


What is the sum of the integers that are both greater than 3 and less than 12?

These numbers cannot include 3 and 12
so $ 4\plus{}5\plus{}6\plus{}7\plus{}8\plus{}9\plus{}10\plus{}11\equal{}60$

You don't have to add them, you could always do $ \frac {(11 \plus{} 4)(11 \minus{} 4 \plus{} 1)}{2} \equal{} \frac {15*8}{2} \equal{} 15*4 \equal{} \boxed{60}$
In this case it might be the same work (I think it is easier), but in general this method is easier.

You don't have to count 4+5+6..... so. So the numbers your adding are 4,5,6,7,8,9,10,11. 11+4= 15, so they're a
group. 10+5=15, so they're a group. And so on, until you have 4,11 and 10,5 and 9,6 and 8,7. Each group equals 15 when added, and there are 4 groups so the answer is 5 times 5, which is 60. :ninja:


inequalities
analytic geometry
trigonometry
modular arithmetic
geometry
geometric transformation
dilation

Let n points ($ n\ge4$)such that distance of any two points is in Z.Prove that minimum is $ \frac{1}{6}$ distances which are multiples of 3

No one can do it?
I think it seems very difficult .Who can solve it?

Assume we have proven it for the base case $ n \equal{} 4$. The simple approach of arguing
[quote]
if we take any $ 4$ of the $ n$ points given, at least one of the $ 6$ distances is divisible by $ 3$, hence the proportion of these distances among the whole must also be at least $ 1/6$
[/quote]
is both pernicious and fallacious. Let us think it in graph terminology. The $ n$ vertices (represented by the points) are connected by edges (where the distance between the two points is divisible by $ 3$). Knowing that any induced subgraph on $ 4$ vertices has at least one edge, estimate the least number $ m$ of edges for the whole graph. Looking at the complement graph, this is tantamount to estimating the largest number of edges for a graph on $ n$ vertices that contains no $ K_4$. But the answer is given by Turan's theorem. We prefer to give a proof based on the Caro-Wei theorem.

[b]Theorem.[/b] For a finite graph $ G \equal{} (V,E)$, denoting by $ \alpha(G)$ the independence number of $ G$ (the largest number of vertices in an induced subgraph with no edges), the following holds $ \displaystyle \alpha(G) \geq \sum_{v \in V} \frac {1} {\deg v \plus{} 1}$.

So in our case we need $ \displaystyle 3 \geq \alpha(G) \geq \sum_{i \equal{} 1}^{n} \frac {1} {\deg v_i \plus{} 1} \geq n\frac {1} {\frac {\sum_{i \equal{} 1}^{n} \deg v_i} {n} \plus{} 1} \equal{} n\frac {1} {\frac {2m} {n} \plus{} 1} \equal{} \frac {n^2} {2m \plus{} n}$, by Jensen's inequality (since $ \frac {1} {x \plus{} 1}$ is a concave function). Therefore $ m \geq \frac {n(n \minus{} 3)} {6} \geq \frac {1} {6} \genfrac {(} {)} {0pt} {} {n} {2} \equal{} \frac {1} {6} \cdot \frac {n(n \minus{} 1)} {2}$ for $ n\geq 5$.

Now for the base case $ n \equal{} 4$, we may choose coordinates such that the four points are $ O(0,0), A(a,0), B(b\cos\beta, b\sin\beta), C(c\cos\gamma, c\sin\gamma)$, with $ a,b,c \in \mathbb{Z}_ \plus{} ^*$. The squares of the six distances are then $ a^2, b^2, c^2, a^2 \plus{} b^2 \minus{} 2ab\cos\beta, a^2 \plus{} c^2 \minus{} 2ac\cos\gamma, b^2 \plus{} c^2 \minus{} 2bc\cos(\beta \minus{} \gamma)$, and I see no other way than a tedious number theoretical analysis, where probably we get to need the angles related to some Pythagorean triangles, where divisibility by $ 3$ could be checked.

For the base case $ n\equal{}4$, by choosing convenient coordinates we can assume the points are $ O(0,0), A(a,0), B(b\cos\beta, b\sin\beta)$ and $ C(c\cos\gamma, c\sin\gamma)$, with the squares of their pairwise distances given by $ a^2, b^2, c^2, r^2 \equal{} a^2\plus{}b^2\minus{}2ab\cos\beta, s^2 \equal{} a^2\plus{}c^2\minus{}2ac\cos\gamma$ and $ t^2 \equal{} b^2\plus{}c^2\minus{}2bc\cos(\beta \minus{}\gamma)$, where $ a,b,c, r, s, t \in \mathbb{Z}_\plus{}^*$.

Assume none is divisible by $ 3$, so $ a^2 \equiv b^2 \equiv c^2 \equiv r^2 \equiv s^2 \equiv t^2 \equiv 1 \pmod{3}$.
We have $ \cos\beta \equal{} \frac {a^2\plus{}b^2\minus{}r^2} {2ab} \equal{} \frac {p} {q}$ with $ (p,q) \equal{} 1$, $ \cos\gamma \equal{} \frac {a^2\plus{}c^2\minus{}s^2} {2ac} \equal{} \frac {u} {v}$ with $ (u,v) \equal{} 1$, so we also have $ p^2 \equiv q^2 \equiv u^2 \equiv v^2 \equiv 1 \pmod{3}$.
Applying a dilation of ratio $ qv \not \equiv 0 \pmod{3}$ to the whole configuration, the points become $ O(0,0), A'(aqv,0), B'(bpv, \pm bv\sqrt{q^2\minus{}p^2})$ and $ C'(cqu, \pm cq\sqrt{v^2\minus{}u^2})$, so the squares of the distances become $ 1, 1, 1, (aqv \minus{} bpv)^2, (aqv \minus{} cqu)^2$ and $ (bpv \minus{} cqu)^2$ modulo $ 3$.

But among the three integer abscissa $ aqv, bpv, cqu$, congruent to $ \pm 1$ modulo $ 3$, there must exist two with their difference divisible by $ 3$, so one of the latter three distances will be $ 0$ modulo $ 3$, contradiction.

[quote="mavropnevma"]For the base case $ n \equal{} 4$, by choosing convenient coordinates we can assume the points are $ O(0,0), A(a,0), B(b\cos\beta, b\sin\beta)$ and $ C(c\cos\gamma, c\sin\gamma)$, with the squares of their pairwise distances given by $ a^2, b^2, c^2, r^2 \equal{} a^2 \plus{} b^2 \minus{} 2ab\cos\beta, s^2 \equal{} a^2 \plus{} c^2 \minus{} 2ac\cos\gamma$ and $ t^2 \equal{} b^2 \plus{} c^2 \minus{} 2bc\cos(\beta \minus{} \gamma)$, where $ a,b,c, r, s, t \in \mathbb{Z}_ \plus{} ^*$.

Assume none is divisible by $ 3$, so $ a^2 \equiv b^2 \equiv c^2 \equiv r^2 \equiv s^2 \equiv t^2 \equiv 1 \pmod{3}$.
We have $ \cos\beta \equal{} \frac {a^2 \plus{} b^2 \minus{} r^2} {2ab} \equal{} \frac {p} {q}$ with $ (p,q) \equal{} 1$, $ \cos\gamma \equal{} \frac {a^2 \plus{} c^2 \minus{} s^2} {2ac} \equal{} \frac {u} {v}$ with $ (u,v) \equal{} 1$, so we also have $ p^2 \equiv q^2 \equiv u^2 \equiv v^2 \equiv 1 \pmod{3}$.
Applying a dilation of ratio $ qv \not \equiv 0 \pmod{3}$ to the whole configuration, the points become $ O(0,0), A'(aqv,0), B'(bpv, \pm bv\sqrt {q^2 \minus{} p^2})$ and $ C'(cqu, \pm cq\sqrt {v^2 \minus{} u^2})$, so the squares of the distances become $ 1, 1, 1, (aqv \minus{} bpv)^2, (aqv \minus{} cqu)^2$ and $ (bpv \minus{} cqu)^2$ modulo $ 3$.

But among the three integer abscissa $ aqv, bpv, cqu$, congruent to $ \pm 1$ modulo $ 3$, there must exist two with their difference divisible by $ 3$, so one of the latter three distances will be $ 0$ modulo $ 3$, contradiction.[/quote]

thanks mavropnevma for your solution
But can u tell me more details in relation of A;B;C and A';B';C'? :blush:
Why do we have A';B';C'?
And why are aqv;bqv;cqv congruent to 1 modulo 3?
I am not sure that if $ a\equiv1$(mod3);$ q\equiv1$(mod3) and $ v\equiv\minus{}1$(mod3)=>$ aqv\equiv1$(mod3) ??? :maybe:

1. The dilation of ratio $ qv$ (a homothety of center $ O$) is only needed because the distances between the points $ O,A,B,C$ are only known to be rational. By doing this homothety, we get the points $ O,A',B',C'$ with integer distances between them.

2. I never said that from $ a, q, v$, congruent to $ \pm 1$ modulo $ 3$, it must follow that $ aqv \equiv 1 \pmod{3}$. If you read the last part, I only said that $ aqv, bpv, cqu$, are congruent to $ \pm 1$ modulo $ 3$. This in turn implies that two of them must have same congruence (pigeonhole principle: three numbers, and only two congruences $ \minus{}1$ and $ 1$).

Can u tell me solution with n>4? :D
I haven't already proved with n>4


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please help me finish proving this problem:

let h be a function from R to R
suppose h'' exists everywhere
show that there exists c in [-1,1] such that h(-1) + h(1) - 2h(0) = h''(c)

so far, applying the/a mean value theorem i know that there exists a in [-1,0] and b in [0,-1] such that h'(a) = h(0) - h(-1) and h'(b) = h(1) - h(0).
next, applying the mean value theorem once more, i know there exists c in [a,b] (and hence in [-1,1]) such that (b - a)h''(c) = h'(b) -h'(a) = h(-1) + h(1) - 2h(0).

that is, the factor of (b - a) is not cool!  it doesn't follow that b - a = 1 if you consider a case like h(x) = x^3.

however, in the case where h(x) = x^3, we know that h''(c)(b - a) = h(-1) + h(1) - 2h(0) = -1 + 1 = 0, so perhaps if we assume h''(c) is not 0, then it follows that (b - a) = 1.  i have tried to find a contradiction by assuming (b - a) is not 1 when h''(c) is not 0, but with no luck.  please help. thanks

Define $ p\equal{}h(1)\minus{}h(\minus{}1)$ and $ q\equal{}h(\minus{}1)\minus{}2h(0)\plus{}h(1).$ Note that $ q$ is the specific quantity we're interested in.

Let $ f(x)\equal{}h(x)\minus{}\left[h(0)\plus{}\frac12\,px\plus{}\frac12\,qx^2\right].$

By explicit calculation, we see that $ f(\minus{}1)\equal{}f(0)\equal{}f(1)\equal{}0.$

Then by Rolle's Theorem, there exist $ a\in (\minus{}1,1)$ and $ b\in (0,1)$ such that $ f'(a)\equal{}f'(b)\equal{}0.$ And then by Rolle's Theorem again, there exists $ c\in(a,b)$ such that $ f''(c)\equal{}0.$

But $ f''(x)\equal{}h''(x)\minus{}q,$ so $ h''(c)\equal{}q$ and we are done.

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This kind of estimate is common in numerical analysis, and numerical analysis proofs quite frequently make use of interpolating polynomials, as I have here.

A different approach:
$ f(t\plus{}1)\minus{}f(t)\equal{}\int_t^{t\plus{}1}f'(x)\,dx$
$ h(1)\minus{}h(0)\minus{}h(0)\plus{}h(\minus{}1)\equal{}\int_0^1 h'(x)\,dx\minus{}\int_{\minus{}1}^0 h'(x)\,dx\equal{} \int_{\minus{}1}^0 h'(x\plus{}1)\minus{}h'(x)\,dx\equal{}\int_{\minus{}1}^0\int_0^1 h''(x\plus{}y)\,dy\,dx$
The desired quantity $ C\equal{}h(1)\minus{}2h(0)\plus{}h(\minus{}1)$ is the average value of $ h''$ over this square.

This can be turned into two applications of the MVT: we first show that $ h'(x\plus{}1)\minus{}h'(x)\equal{}C$ for some $ x$, and then that $ h''(x\plus{}y)\equal{}C$ for some $ y$.

and why would my professor believe i did that on my own? haha--actually he doesn't care about anything, so... i think he hates life, lol

You do realize that Kent is a professor and that I've taught a lot of calculus myself (mostly as a TA)? We're trying to teach- to show you what the right ideas are, so that you can put them together. Try to understand how each of us got around the difficulty you noted, and apply one of those methods to build the solution yourself. You'll understand it better that way.
Also, I find it very unlikely that your professor "doesn't care about anything". We may not have time to deal with individual issues during class, but there is more to teaching than just the actual class hours.