if  $\lambda$  is a real root of $P(x) = x^5 - x^3 + x - 2$ ,then prove that
$[ \lambda ^6] = 3$ where the $[a]$ denotes the greatest integer function.

Since P'(x) = 5x 4  - 3x 2  + 1 > 0 (consider it as  a quadratic expression in x 2 , and this expression is positive), we deduce that P is an increasing function. Thus, there is only one solution of the equation P(x) = 0 (there is one from the intermediate values theorem), say a instead of  \lambda  which is not very easy to read.

Now, P( 2 1/3 ) = 10*2 1/3 - 2 + 2 1/3 - 2 > 10 - 4 > 0.
Moreover P(0) = -2.
It follows that 0 < a < 2 1/3 and then 0 < a 6 < 4.

In another hand, P(x) = -2 + x(1 + x 6 )/(1+x 2 ). Thus :
a 6 = -1 + 2(a + 1/a)  \geq -1 + 2*2 = 3.
Thus 3 < a 6  < 4 and we are done.

Pierre.

I did the same to show P(21/3)>0 and for a6 >3 I did this:

(a-1)(a4+a3+1)=1 and (a-1)(a5+a4+a3+a2+a+1)=a6-1>=2 because we can show that a>1 and a5+a2>=a4+a3.

Nice problem!

my solution is nearly the same ,but it involvs more bull work (so typical of me)
what i proved was that if  $|x| \geq 1$ and $x^5-x^3 + x - 1 = a$ then
 $x^6 - 1  \geq  2a$. (but i first proved the bounds on x to be (1.2<x<1.25) :D which follows
$x^6-1 =(x-1)(x^5+x^4+x^4+x^2+x+1)$ but $(x-1)= [a/(x^4+x^3+1)]$
\[ x^6-1 = a \frac {x^5+x^4+x^4+x^2+x+1}{x^4+x^3+1} \]
that the term inside the square brackets is greater than 2 is equivalent to proving $(x-1)[(x^2-1)^2 + x^2]  \geq  0$ which is true as $x>1$ by assumption.hence $x^6-1 \geq  2a$ (and $a=1$ for this particular problem)

the latex command for the greek letter lambda $\lambda$


Hi.

 I have this problem on projectile motion. I have attached the problem as a pdf file. I'm wondering if someone can check my result. The problem is posed and I have included a diagram.

Many thanks in advance.

are the bodies identical ?yes then look here is my answer.
they collide when their vertical velocity become zero or we can say when they reach d highest point  .
till that point horizontal distance =$ \frac {U^2sin2\alpha}{2g}$
after that the new velocity of the system (body A +body B) = $ U\frac{1}{2} [cos\alpha - \frac {sin(\alpha)cos(\beta)}{sin\beta}]$
the time of duration the system would be in air after collision =$ \frac {Usin(\alpha)}{g}$
so the total horizontal dis = ${ \frac {U^2sin2\alpha}{2g} + \frac {\frac{U^2}{2} sin(\alpha)}g}( cos(\alpha) - \frac {sin(\alpha)cos(\beta)}{sin(\beta)})$
=$ \frac {U^2sin2\alpha}{2g} + \frac {U^2sin(\alpha)}{2g sin(\beta)}(sin(\beta - \alpha))$
thank u
Pardon if i have done some mistake .


Solve in the set of integers the equation :

   $ x^8\plus{}2^{2^x\plus{}2} \equal{} p$

 where $ p$ is a prime positive integer.

we can easely prove that x^4+4y^4 is not prime except x=y=1 since it can be factorized. so x cannot be greater than 1.


function
calculus
derivative
calculus computations

given a Cost function C(x)=30x , and a Revenue function R(x)=-3x^2  +80

Find the quantity that leads to the maximum profit.
 i solved and found x=-5.
is this question right???

Profit function will be
$ P(x)\equal{}R(x)\minus{}C(x)$

$ P(x)\equal{}\minus{}3x^2\minus{}30x\minus{}80$

$ P'(x)\equal{}0$ gives $ x\equal{}\minus{}5$
Check that $ P''(\minus{}5)<0$

Thus profit is maximised.

thanks. but x the number of units that are produced. how can we explain the negative sign??????

please someone answer me: can x be negative??

x is the number of sold units . how can x be negative

So it can't be negative. When you're looking for a maximum or minimum, the zeros of the derivative aren't the only points of interest.


function
trigonometry
real analysis
real analysis unsolved

Find a bijection f:[0,1]-->R.I need exactly a function like that,please dont only prove there exists f.

[quote="Ibrogimov"]$ y \equal{} \tan(\pi x \minus{} \frac {\pi}{2})$[/quote]What is $ f(0)$?

[quote="jmerry"]The closed interval is hard- you can't do it with continuous functions, or even piecewise continuous functions with finitely many pieces. Your best bet is to construct a piecewise continuous function that maps the interval with infinitely many points removed to the real line with infinitely many points removed, and then construct a separate bijection on the two countable sets that remain.[/quote]

could you show an example?I dont understand :(

The key is to cunningly remove the two problem points (0 and 1) by shifting them into the open interval (0,1). So define f as
\[ f(x) = \{\begin{array}{ll}tan(\pi{x} - \frac {\pi}{2}) & x \in (0,1), \frac {1}{x} \notin \mathbb{N} \\
tan(\frac {\pi}{n + 2} - \frac {\pi}{2}) & \frac {1}{x} \in \mathbb{N} \\
0 & x = 0 \end{array}
\]
Essentially, what's happened is that you first biject [0,1] to (0,1), and then compose this bijection with the bijection from (0,1) to $ \mathbb{R}$ given above.

to Token:  $ f(x)\equal{} tan(\frac {\pi}{n \plus{} 2} \minus{} \frac {\pi}{2})$  with     $ \frac {1}{x} \in \mathbb{N}$ [/quote]

where doez $ n\plus{}2$ come from??

Here is a solution bijection f:[0,1]-->(0,1)
$ f(0)\equal{}\frac{1}{2} \\ f(\frac{1}{2^n})\equal{}\frac{1}{2^{n\plus{}1}} ,n \in N^{*} \\
f(\frac{1}{3^n})\equal{}\frac{1}{3^{n\plus{}1}} ,n \in N \\
\text{the rest} f(x)\equal{}x$


Given a circle with center $O$ and two exterior points $A$ and $B$, construct a diameter $PQ$ such that $AP=BQ$.

See the attached diagram.

By the conditions, $AP=BQ$ and $PO=OQ$. If we construct $A'$ such that $O$ is the midpoint of $AA'$ (that is, $A'$ is centrally symmetric to $A$ with respect to $O$, or $A'=\mathcal{S}_{O}(A)$), then $\triangle AOP\cong\triangle A'OQ$, giving $A'Q=AP=BQ$. That, however, means that $Q$ lies on the perpendicular bisector of $A'B$, and now the construction is clear.

Depending on the mutual position of the given circle and the bisector of $A'B$, we can have zero, one or two solutions for the desired diameter.


ceiling function
logarithms
pigeonhole principle
function
LaTeX

Let $ S$ be a subset of $ \{1, 2, 3, ... , N\}$such that no two subsets of S have the same sum. What is the largest possible sum for $ S$ if
a) $ N\equal{}50$
b)$ N\equal{}200$

Well technically, $ \{1,2,3,\ldots,N\}$ is a subset of itself.  So that has a sum
$ \frac{N(N\plus{}1)}{2}$

[quote="JRav"]Well technically, $ \{1,2,3,\ldots,N\}$ is a subset of itself.  So that has a sum
$ \frac {N(N \plus{} 1)}{2}$[/quote]
the question says that no two subsets of S have the same sum, we need to apply this constraint

[hide="Partial result"]

Let $ n$ be the maximum number of elements of $ S$.

Then $ n$ can never be greater than $ \lceil \log_2 {(\frac {N(N \plus{} 1)}{2} \plus{} 1)} \rceil$

[b]Proof:[/b]

If $ S$ has $ n$ elements, than it has exactly $ 2^n$ different subsets.

The sum of any subset can not be smaller than $ 0$ and can't be bigger than $ 1 \plus{} 2 \plus{} 3... \plus{} N \equal{} \frac {N(N \plus{} 1)}{2}$

So the maximium number of substes is at most $ \frac {N(N \plus{} 1)}{2} \plus{} 1$.

(If there would me more subsets, than at least two would (by the Pigenhole priniciple) have the same sum).

So

$ 2^n \le \frac {N(N \plus{} 1)}{2} \plus{} 1$.

$ n \le \log_2 {(\frac {N(N \plus{} 1)}{2} \plus{} 1)}$

Since $ n$ must be an integer

$ n \le \lceil \log_2 {(\frac {N(N \plus{} 1)}{2} \plus{} 1)} \rceil$

$ \blacksquare$

You can aplay this result to lower the bound even lower.
 [/hide]

Wichking, you should be using floor, not ceiling.  Also, rather than iterating, we can do the following:  a set with $ n$ elements has $ 2^n$ subsets.  The sums of those subsets are bounded between $ 0$ and $ n \cdot N$.  Thus $ 2^n \leq n \cdot N \plus{} 1$ and $ N \geq \frac {2^n \minus{} 1}{n}$.  This doesn't have a nice inverse function, but it's easy to see that for $ N \equal{} 50$ we must have $ n < 9$ and for $ N \equal{} 200$ we must have $ n < 12$.  We can take a slighly tighter estimate than $ n \cdot N$ for the maximal sum and maybe reduce these bounds a little bit.

An easy lower bound on the answer is $ 3N \minus{} 3$ when $ N \geq 4$, by taking the largest three elements.  It seems unlikely that this is optimal, though -- in particular, for $ N$ slighly larger we can add 3 to this set without causing any conflicts.  Note also that a larger set doesn't necessarily lead to a larger sum -- taking all the powers of 2 less than $ N$ gives a smaller lower bound (beween $ N$ and $ 2N$) than the bound I gave.  In particular, taking a set of maximal size is definitely not guaranteed to give us a maximal sum.

This seems rather hard for the intermediate forum -- I would have placed it in preolympiad.

JBL I wasn't talking about the [b]maximal sum[/b] I was talking about the [b]maximal number of elemnts[/b] such set can have. I hope that clears some things ,if i wasn' clear enough before... :)

Yes, I know -- I've edited several times now, so you should reread my comment :)

Also, use "\log" instead of "log" to make your latex prettier.

[quote]
Yes, I know -- I've edited several times now, so you should reread my comment [/quote]

I did  :) ... and i agree :)

[quote]Also, use "\log" instead of "log" to make your latex prettier.[/quote]

Thanks :)


[quote="SplashD"]Everybody says that I don't have a quote button next to my posts but nobody else seems to have that problem, why?[/quote]

Do you mean like pressing the quote buttom to do what I just did?

Somehow, your quote button got stuck under the hide tag.

SplashD maybe it would be a good idea to remove that hide tag for now from your signature :)

in ML style it even eats your post date.

it seems that the problem for nested hide tags has expanded its territory to single hide tags too. :P


analytic geometry
geometry
trapezoid
rectangle
perimeter
function
conics

Basic background: I'm a sophomore, and currently in precalculus struggling, especially with coordinate proofs and distance of point to a line or something like that. I was never good at math, but especially everything related to geometry I seem to suck the most.

Anyways, here are some samples of problems in my textbook:

1) Prove that the length of the line joining the midpoints of two sides of a triangle is equal to one-half the length of the third side.

2) An isosceles trapezoid is a trapezoid with legs of equal length. Prove that the diagonals of an isosceles trapezoid are equal in length.

3) Prove that the sum of the squares of the lengths of the three medians of any triangle is equal to three-fourths the sum of the squares of the lengths of the three sides.

And then there's like a word problem, such as: A road is to be constructed to pass through a small mountain. A two-dimensional model for the shape of the mountain is the curve with equation y=24-x^2. A rectangular passageway is to be made through the center of the mountain. Let the x-axis represent the road. Then use a two-dimensional model in which the top vertices of the rectangle touch the curve and the bottom vertices are on the x-axis to find the maximum perimeter of the triangle.

Even if it's just help with 1 problem, I appreciate. Like I said, I'm not talented in math and I don't have high standards of myself in math - I'm below average and I'd be happy with like a B- C+, I guess. It'd help to actually understand at least [b]some[/b] stuff, though.

Thanks for any help.

Ok.  I will help you with number 1.

Any triangle can have three points that are $ (0,0)$, $ (a,0)$, and $ (b,c)$.  Whenever you do a proof with triangles, you should always use this unless you have more information.

Ok, so we have to find the midpoint of $ (0,0)$ and $ (b,c)$.  Clearly, this is $ (\frac{b}{2}, \frac{c}{2})$.

We also have to find the midpoint of $ (a,0)$ and $ (b,c)$.  Clearly, this is $ (\frac{a\plus{}b}{2}, \frac{c}{2})$.

Then we have to find the distances of the $ 2$ lines that we have.

The distance from $ (0,0)$ to $ (a,0)$ is obviously $ a$. (This is that third side.)

The distance from $ (\frac{b}{2}, \frac{c}{2})$ and $ (\frac{a\plus{}b}{2}, \frac{c}{2})$ is $ \frac{a}{2}$.  Thus, we can clearly see that this length is half of the third side.

2).

Draw a picture. That always helps.

[asy]size(200);
draw((0,0)--(10,0)--(8,4)--(2,4)--cycle);
draw((0,0)--(8,4));
draw((10,0)--(2,4));
label("$A$",(0,0),W);
label("$B$",(10,0),E);
label("$C$",(8,4),E);
label("$D$",(2,4),W);[/asy]

If $ \triangle{ABD}\cong\triangle{ABC}$ then $ \overline{AC}\cong\overline{BD}$. That should be enough to start off with.

Coordinate proofs are cumbersome and can turn into confusing letter soup. Grab some paper (grid paper if possible), and start sketching. Label the important points, with their coordinates. That will help you “see” the problems, and keep track of what each letter means.

Ernie showed you how to work problem 1. I’d use a similar strategy for problems 2 and 3. For convenience, I would make $ (\minus{}a,0)$ and $ (a,0)$ the coordinates of the ends of the longer base of the trapezoid. The other two vertices would be $ (x, c)$ and $ (b, c)$. I’d specify that $ a$ and $ c$ are positive (not really a restriction, but a question of personal preference). I would also specify that $ x<b$ making $ (b, c)$ a vertex adjacent to $ (a,0)$. Can you calculate the lengths of the legs now (as functions of $ a, b, c$, and $ x)$? Can you prove that $ x\equal{}\minus{}b$? Can you calculate the length of the diagonals?

I believe I know what the wording of your word problem really meant to mean. (I just don't see any practical application for it).
[quote="CoffeeCandy128"]A road is to be constructed to pass through a small mountain. A two-dimensional model for the shape of the mountain is the curve with equation y=24-x^2. A rectangular passageway is to be made through the center of the mountain. Let the x-axis represent the road. Then use a two-dimensional model in which the top vertices of the rectangle touch the curve and the bottom vertices are on the x-axis to find the maximum perimeter of the triangle.[/quote]The word problem is meant to make you graph the parabola for the equation they gave you. It represents the mountain (in cross-section). It has a vertex at $ (0,24)$ and intersects the x-axis at $ x\equal{}2\sqrt6$, and $ x\equal{}\minus{}2\sqrt6$. The surface of the road is represented by the x-axis. The roof of the tunnel is represented by the line $ y\equal{}b$, where $ b$ is the height of the tunnel. The ends of the tunnel’s roof are at $ (a, b)$ and $ (\minus{}a, b)$. The rectangle in the problem is the one with vertices at $ (\minus{}a, 0), (a, 0), (\minus{}a, b)$ and $ (a, b)$. It represents the tunnel (in cross section). They must want you to find the maximum perimeter of that [b]rectangle[/b] (there is no [b]triangle[/b] anywhere).
Can you calculate the length of the roof and the height of the tunnel as a function of $ a$?
Can you calculate the perimeter $ (P)$ as a function of $ a$?
$ (P)$ as a function of $ a$ is a parabola that opens down, so its vertex is a maximum.
Can you calculate the coordinates ($ a$, and $ P$ values) of the vertex/maximum?
[hide="solution to odd problem"] The length of the roof is $ 2a$.
The height of the tunnel is $ b\equal{}24\minus{}a^2$.
The perimeter is $ P\equal{}\minus{}2a^2\plus{}4a\plus{}48$, which is a parabola with a vetex at $ x\equal{}1, P\equal{}50$.
The maximum perimeter of the rectangle is 50.[/hide]
Disclaimer: I often make mistakes. If you (meaning anyone) see one (or three), point it/them out for me.


In a right triangle $ ABC$, point $ F$ is the perpendicular foot of $ C$ on hypotenuse $ AB$. Let point $ Q$ on $ AB$ be the perpendicular foot of a certain point $ P$ on leg $ AC$. The line through $ B$ which is perpendicular to $ BP$ intersects line $ CF$ at point $ R$. Show that the intersection of circles with diameters $ AC$ and $ FQ$, different from $ F$, lies in the interior of segment $ PR$.

Denote by $ O$ the point which is on the same part of $ AB$ like the point $ C$. Denote by $ S$ the foot of the altitude from $ F$ to $ OQ$.

It is enough to prove that $ P\minus{}S\minus{}R$ lie on the same line.

Now $ O(0,0),C(a,0),A(0,b)$ and $ F(a,b)$. $ W.L.O.G.$ let's take $ Q(1,b)$

$ \Rightarrow B(\frac {a^2\plus{}b^2} {a},b), P(1,b \cdot \frac {a\minus{}1} {a}),S(\frac {a\plus{}b^2} {1\plus{}b^2},b \cdot \frac {a\plus{}b^2} {1\plus{}b^2})$ and $ R(a,\frac {b(a^2\plus{}b^2)} {a})$

The conclusion results easily.

[quote="moldovan"]Denote by $ O$ the point which is on the same part of $ AB$ like the point $ C$. [/quote]

What do you mean?


Solve systems of equation $ \begin {cases} x^2-4\sqrt{3x-2}+6=y\\y^2-4\sqrt{3y-2}+6=x\end{case}$

[hide="real solution"]Since the two equations are inverses of each other, all you need to do to find the real solution is find where they intersect the line $ y\equal{}x$. In other words, find the fixed point(s).[/hide]

[quote="t0rajir0u"]Unfortunately, it's not generally true that the graph of a function and its inverse intersect only on the line $ y \equal{} x$.[/quote]
A concrete example: $ y \equal{} \frac {1}{x}$...

I think that since they are inverses, if you set $ f(x)\equal{}x^2\minus{}4\sqrt{3x\minus{}2}\plus{}6$, then $ f(f(x))\equal{}x$, and that might be the way to solve it, but it might also make a really long ugly equation.

$ f(x)\equal{}x^2\plus{}6\minus{}4\sqrt{3x\minus{}2}$, $ f(x)\ge x$ is true...

$ x^2\minus{}x\plus{}6\ge 4\sqrt{3x\minus{}2}$

$ (x^2\minus{}x\plus{}6)^2\ge 16(3x\minus{}2)$

$ (x\minus{}2)^2(x^2\plus{}2x\plus{}17)\ge 0$ woo

Then $ x\equal{}f(f(x))\ge f(x)\ge x$ equality iff $ x\equal{}2$, we see that this does work.

Wow, the best ideas. thanks a lot [b]Altheman[/b]


Can someone help me with calculation of moments of inertia? The moments of inertia for a rectangular box and a flat rectangular "board" through both their centers. I don't know where to start with the integration!

:D Thanks!!

I've removed this post, and replaced it again. I thought there was an error in it, but it checked out OK.

The attachment shows how to do the integration for the 1st MOI.

If there's anything not clear, please post back.

There's a better illustration of [url=http://www.mame.mu.oz.au/~mcg/436-121/lecture_07.pdf]things here[/url], and a shorter treatment of the working out.


[size=150]$ x,y,z > 0$,prove that:

(1)
$ \frac {2(x \plus{} y)^2 \minus{} (y \plus{} z)^2}{z \plus{} x} \plus{} \frac {2(y \plus{} z)^2 \minus{} (z \plus{} x)^2}{x \plus{} y} \plus{} \frac {2(z \plus{} x)^2 \minus{} (x \plus{} y)^2}{y \plus{} z} \geq 2(x \plus{} y \plus{} z).$

(2)
$ \frac {3}{8}\geq \frac {x(x \plus{} y)}{(2x \plus{} y \plus{} z)^2} \plus{} \frac {y(y \plus{} z)}{(2y \plus{} z \plus{} x)^2} \plus{} \frac {z(z \plus{} x)}{(2z \plus{} x \plus{} y)^2}$.[/size]
BQ

[quote="xzlbq"][size=150]$ x,y,z > 0$,prove that:

(1)
$ \frac {2(x \plus{} y)^2 \minus{} (y \plus{} z)^2}{z \plus{} x} \plus{} \frac {2(y \plus{} z)^2 \minus{} (z \plus{} x)^2}{x \plus{} y} \plus{}$ $ \frac {2(z \plus{} x)^2 \minus{} (x \plus{} y)^2}{y \plus{} z} \geq 2(x \plus{} y \plus{} z).$
BQ[/quote]

assume  $ x \equal{} max{(x,y,z)}$

$ (1) \Longleftrightarrow (2x^2 \plus{} (6y \plus{} 2z)x \plus{} 4y^2 \plus{} 2yz)(x \minus{} y)(x \minus{} z)$ $ \plus{} ((4y \plus{} 4z)x \plus{} 2(y \plus{} z)^2)(y \minus{} z)^2\geq 0$


The 3rd and 5th terms of an arithmetic sequence are 17 and 39, respectively. What is the 7th term of the same sequence?

The difference between the 3rd/5th terms and 5th/7th terms is equal in an arithmetic sequence.

$ 39 \minus{} 17 \equal{} 22$   find the difference

$ 39 \plus{} 22 \equal{} 61$   add to the 5th term to obtain the 7th


symmetry
arithmetic sequence
6th edition

How was this problem? Did you solve it?

I didn't try to solve it, I was kind of lazy these weeks. (and also had many tests)

Look at the differences between the $x_i$, there are $n(n-1)/2$, all integers between $1$ and $n(n-1)/2$.

If two of them are equal, then there are three $x_i$ in arithmetic progression, or $x_i+x_j=x_k+x_l$ with$i$,$j$,$k$,$l$ distinct.

If they are all different then all differences appear, and by trying to construct a working example (starting with a difference of $n(n-1)/2$ and going on), we find a contradiction for $n \geq 5$.

How do you derive this contradiction?  It seems to me that this was the meat of the question.
Jack

You will get that the differences x(i+1)-x(i) where x(n),x(n-1)... are in ascending order are 1,2,3..n-1 so the numbers will be x(1),x(1)+1,x(1)+3,x(1)+6... in some order and x(1),x(1)+3 and x(1)+6 are in arithmetic progression.Contradiction

Your solution doesn't seem correct, Makaveli : there is a contradiction only if $n \geq 5$, and you don't use it.

Here comes an extract from my (ugly) solution :

(I've named $X=\{x_i\}$, $E=\{|x_i-x_j|\}$, and $N=\frac{n(n-1)}{2}$.)

...

We thus have $N \in E$, so for some $1\leq a,b \leq n$, $x_a = x_b + N$.
By subtracting $x_b$ to all $x_i$, we can reduce to the case $x_b=0$
and $x_a=N$. Since $|x_i - x_b|$ and $|x_i - x_a|$ are at most $N$,
$X\subset [0,N]\cap\mathbb{Z}$.

$N-1 \in E$, so either $1$ or $N-1$ is in $X$. By symmetry reasons, both
cases are equivalent, so we can study only the first one : $\{0,1,N\}\subset X$.

$N-2 \in E$, and $N-2 > 1$, so either $2$, $N-2$ or $N-1$ is in $X$.
$2$ and $N-1$ cannot be in $X$ because $2-1 = 1-0$ and $N-(N-1) = 1-0$,
so $\{0,1,N-2,N\}\subset X$.

$N-4 \in E$, and $N-4 > 2$, so $2$, $4$, $N-4$ or $N-3$ is in $X$. We have $2-1=1-0$,
$(N-2)-(N-4) = N-(N-2)$, and $(N-2)-(N-3) = 1-0$, so $4 \in X$.

We thus have $\{0,1,4,N-2,N\}\subset X$. The differences of numbers in $\{0,1,4,N-2,N\}$
are $\{1,2,3,4,N-6,N-4,N-3,N-2,N-1,N\}$. We have $n\geq 5$, so $N\geq 10$,
and $N-5 > 4$, thus $N-5$ doesn't appear yet in the set of all differences of $x_i$.

$N-5 \in E$, so $3$,$5$,$N-5$,$N-4$ or $N-1$ is in $X$. Since $4-3=1-0$, $5-4=1-0$,
$(N-2)-(N-5) = 4-1$, $(N-2)-(N-4)=N-(N-2)$ and $N-(N-1)=1-0$, we get a contradiction.

Guys I only used the problem in the $IMO2003$ and get to the contradiction  :lol:
Here it
 $\left(\sum_{i,j=1}^{n}|x_i-x_j|\right)^2\le\frac{2(n^2-1)}{3}\sum_{i,j=1}^{n}(x_i-x_j)^2.$  ;)

What is my mistake then?Please tell me!Who says the fact n>=5 is really necessary?I used n>3

Have I a wrong impression or toumaf's solution is only for n>5?

Infact in the set 0,1,4,N-2,N there are 5 elements but the equation we have to prove is inccorect...

(sorry for my english)

I have more powerful solution using $Sidon-sequence$

[color=red]We even don't need the condition "no 3 of them can be in arithmetic progression" for $n\geq 13$[/color]

Let's use $S-sequence$ which means sequences of nonnegative integers with the property that

all pairwise sums of elements are distinct.

Denote $S(n)$ by the maximal number of elements of an $S-sequence$ with elements not greater than $n$

Then we get $\sqrt{n}- n^{\frac{11}{40}}< S(n)< \sqrt{n} + n^{\frac{1}{4}} +1$.

It is obvious that adding same value to $x_i$s is no problem. So we can make $x_i$ is the sequence of

nonegative integers staring at $0$.

cuz the maximal range of the sequence is $\frac{n(n-1)}{2}$ by the condition $2|x_i - x_j|\leq n(n-1)$,

We get $S(t)< \sqrt{t} + t^{\frac{1}{4}} +1$ for $t=\frac{n(n-1)}{2}$.

But $n>\sqrt{t} + t^{\frac{1}{4}} +1$ for $n\geq 13$, we get contradiction.

and for $n\leq 13$ it is trivial.

$Q.E.D.$

[quote="Chang Woo-JIn"]
Then we get $\sqrt{n}- n^{\frac{11}{40}}< S(n)< \sqrt{n} + n^{\frac{1}{4}} +1$.
[/quote]
How to prove this one? :roll:


a lot of people here have a lot of experience with Olympiad questions.  obviously more experience the better you are at.  i do not feel comfortable solving olympiad questions.  how long is it before you guys feel comfortable doing olympiad questions?

[quote="Valentin Vornicu"]took me 2 years to get first prize at the Romanian National Olympiad back in 1998, a few moths to score Gold at the JBMO, and a couple more years to go to the IMO / BMO - one more to get a medal, and probably another one (but i've got too old already :D) to score gold.

Some people skip the first prize parts, and get really quick to the IMO Gold stuff :D - others just stuck with the Nationals - anyway don't worry about it, as long as you like what you do, it won't matter that much what you did in the exam  :D[/quote]

what's JBMO?

Junior Balkan MAthematical Olympiad - it's been active for 7 years now - it's sort of Balkan Olympiad but only with students younger than 15 1/2 years at the moment of the contest. At the downloads section I have put up some of the subjects from past years.

Without any preparation or so I got a second prize at the Flanders Mathematics Olympiad for JUNIORS in 2002. After that I started to do olympiad problems - at first I did only very simple multiple choice questions, then I did Flemish final questions, I got interested in other national competitions (I started with the Dutch, after that the British ...). And now, one year later, I'm mainly doing IMO - level questions.
This year I won first prize in the Flanders Mathematics Olympiad, and I got selected for the IMO. I did allmost all of my preparation on my own. You know, Belgium is divided in two parts : Flanders (Dutchspeaking) and Wallonia (Frenchspeaking). Now the Wallonian guys have +/- 7 preparation weekends each year + correspondence courses and selection tests and so on. The Flemish guys have only one weekend on their own, and this year was the first year in which Flemish & Wallonian guys did a "common weekend". So I got two weekends preparation for the IMO. In general Wallonians perform better then Flemish guys at the IMo because of their more extensive preparation. (Flemish guys are not sponsored by the authority, Wallonian guys are sponsored).
I did this year's IMO and I got 10 points (HM). That's not brilliant to Romanian norms, but it's quite good for a Flemish guy. For comparison :
Wallonian results : 22 - 17 - 8
Flemish results : 10 - 8 - 5
So I mean that the more you train, the better you are. Now I'm doing olympiad questions for one year and I'm still pretty weak ... But I enjoy problem solving and that's the most important thing ! :)

I hope my story is a bit helpful & interesting for you ...

I do think i'm extremely extremely weak in olympiad questions...

On non-IMO contests I was in a contest for primary school students for the whole China in 1996 which I think I take several months to get used to it, then HKMO (not IMO types, just fill-in-the-blanks) in 1998 after 3 months of training, my team coming 17th, 1st, 2nd on 1998-2000.
On IMO style problems, it took me far more time. When I first entered the training team in 1998 I don't even know what is geometry and don't even know the fact that angles on an arc are always the same because it's not teached in school. I did only number theory and alegbra problems, getting very low scores in tests. For next year it's better because I finally learn the basic geometry and got selected for IMO 2000. It was the time I feel I got used to IMO-style problems, getting a silver. I managed to get a gold on 2002 although it's in fact 7-7-0-7-7-1. For 2003 I was so busy with public exams that I almost attended no training sessions and I come down to 7-7-3-0-1-1, and I was very depressed. After entering university I haven't even done a IMO-style problem...

[quote="Charlie"]... I was very depressed. After entering university I haven't even done a IMO-style problem...[/quote]

When you are in university there exist IMC International Mathematics
Competition
http://www.mathlinks.ro/phpBB/viewtopic.php?t=541

Maybe next summer you can try.
Be optimistic

Moubinool Omarjee

Well, what I meant was just that I've no time to do them... Then again, no university in Hong Kong is sending an IMC team  :(

you can participate as an individual, also ... all that you need is the transportation money ... and food & stuff, so it's about 1,500 USD.

When and where will it be hosted on 2004? Hope to go by rail and stay at YMCA to keep the budget at around USD 600...

Bulgaria, so it's kind of far for you Charlie. Go to the home page of IMC, I don't remember it, but try Google, and get it touch with John E. Jane. He is in charge with the organization and participation problems. He can't pay your trip, but he might give you better info on the amount of cash you need. In 2003 there participated a group from Iran, and it was held in Romania. (if you find Bulgaria on a map, look up to find Romania - it's in Eastern Europe)
What is YMCA?I know a song about this one, but I don't think that is it.

I think I'm pretty experienced, but Olympiads are a whole new topic
too hard :(


$\text{Describe the graph of the following equation: } 6x^2 + 5y^2 + 13xy + x - 3y -2 = 0$

[quote="shobber"][hide]
$13^2-4\cdot 5\cdot 6=169-120=49>0$
So it is a hyperbola.[/hide][/quote]

Wrong. !!!  ;)

[quote="shobber"]well from what i know, the graph of a second degree equation is always a conic.[/quote]

you may miss in some concept.  ;)

[quote="shobber"]well from what i know, the graph of a second degree equation is always a conic.[/quote]
There are some degenerate cases like two lines, or point, or even empty set ;)

Myth and svrsk have the correct approach

that was a cool trick you used there svrsk  :)


Αν για τους πραγματικους αριθμους $ x, y, z$ ισχυει οτι:

$ \frac {(x \minus{} y)^3 \plus{} (z \minus{} 2x)^3}{(x \plus{} y \minus{} z)^3} \geq \minus{} 1$

τοτε να αποδειξετε οτι:

$ {(\frac {x \minus{} y}{y \plus{} z} \plus{} \frac {y \plus{} z}{z \minus{} 2x})}^2 \plus{} {(\frac {y \plus{} z}{x \minus{} y} \plus{} \frac {z \minus{} 2x}{y \plus{} z})}^2 \geq 8$

[quote="NickNafplio"]Αν για τους πραγματικους αριθμους $ x, y, z$ ισχυει οτι:

$ \frac {(x \minus{} y)^3 \plus{} (z \minus{} 2x)^3}{(x \plus{} y \minus{} z)^3} \geq \minus{} 1$

τοτε να αποδειξετε οτι:

$ \underbrace{{(\frac {x \minus{} y}{y \plus{} z} \plus{} \frac {y \plus{} z}{z \minus{} 2x})}^2 \plus{} {(\frac {y \plus{} z}{x \minus{} y} \plus{} \frac {z \minus{} 2x}{y \plus{} z})}^2 }_{K}\geq 8$[/quote]

Είναι $ K\geq 4\plus{}2(\frac{x\minus{}y}{z\minus{}2x}\plus{}\frac{z\minus{}2x}{x\minus{}y})$, αλλά από τη δοθείσα σχέση αν θέσουμε $ A\equal{}x\minus{}y$ και $ B\equal{}z\minus{}2x$ έχουμε $ \frac{A^3\plus{}B^3}{(A\plus{}B)^3}\leq 1$ και ισοδύναμα καταλήγουμε $ A\cdot B\geq 0\Longrightarrow \frac{A}{B}\geq 0$ άρα  $ K\geq 8$.

$ qed$

Σωστος Nick
Αλλα μου βγηκε ευκολοτερη απ οσο περιμενα οποτε μετα θα ανεβασω μια αλλη λιγο πιο δυσκολη

η δοθεισα σχεση ειναι ισοδυναμη με:
$ (x \minus{} y)(z \minus{} 2x)(z \minus{} (x \plus{} y)) > \equal{} 0$αυτο ισχυει οταν x>=y,z>=2x,z>=x+y η το ακριβως αντιστροφο.
η ανισοτητα μας ομως που θελουμε να αποδειξουμε ειναι $ > \equal{} 4(\frac {x \minus{} y}{z \minus{} 2x} \plus{} \frac {z \minus{} 2x}{x \minus{} y})$(1)
διακρινω περιπτωσεις:
1)αν ισχυει
$ x > \equal{} y,z > \equal{} 2x,z > \equal{} x \plus{} y \minus{} \minus{} > x \minus{} y > \equal{} 0,z \minus{} 2x > \equal{} 0 \minus{} \minus{} > (z \minus{} 2x),(x \minus{} y)$omosimoi
τοτε εχουμε$ (1) > \equal{} 4*2 \equal{} 8$(στηριζομενoς στην γνωστη $ a/b \plus{} b/a > \equal{} 2$(gia omosimous a,b)

2)αν $ x < \equal{} y,z < \equal{} 2x \minus{} \minus{} > (x \minus{} y),(z \minus{} 2x)$pali omosimoi ara $ (1) > \equal{} 8$και παλι... :)

Σωστος και ο Αθηναιος

Η Παρακατω ασκηση ειναι σαν μια γενικευση αυτης που εβαλα πιο πανω (δεν κανω καινουργιο τοπικ αφου οι δυο ασκησεις δεν διαφερουν και πολυ)

Ασκηση:

Εστω μια Αριθμιτικη Προοδος πραγματικων αριθμων $ (a_n)$.
Εστω ακομα οτι ισχυει: $ {a_i}^3 \geq {a_{i \minus{} 1}}^3 \plus{} {(a_{j \plus{} 1} \minus{} a_j)}^3$
για $ i \equal{} 2, 3, 4, ..., n$ και $ j \equal{} 1, 2, 3, ..., n \minus{} 1$

Να αποδειξετε οτι:

$ {( \frac {a_1}{a_2} \plus{} \frac {a_2}{a_3} \plus{} ... \plus{} \frac {a_{k \minus{} 1}}{a_k})}^2 \plus{} {(\frac {a_k}{a_{k \minus{} 1}} \plus{} \frac {a_{k \minus{} 1}}{a_{k \minus{} 2}} \plus{} ... \plus{} \frac {a_2}{a_1})}^2 \geq 2k^2 \minus{} 4k \plus{} 2$

για καθε φυσικο αριθμο $ k$ με $ 2 \leq k \leq n$.

Ποτε ισχυει η ισοτητα;

[quote="NickNafplio"]
Εστω μια Αριθμιτικη Προοδος πραγματικων αριθμων $ (a_n)$.
Εστω ακομα οτι ισχυει: $ {a_i}^3 \geq {a_{i \minus{} 1}}^3 \plus{} {(a_{j \plus{} 1} \minus{} a_j)}^3$
για $ i \equal{} 2, 3, 4, ..., n$ και $ j \equal{} 1, 2, 3, ..., n \minus{} 1$

Να αποδειξετε οτι:

$ {( \frac {a_1}{a_2} \plus{} \frac {a_2}{a_3} \plus{} ... \plus{} \frac {a_{k \minus{} 1}}{a_k})}^2 \plus{} {(\frac {a_k}{a_{k \minus{} 1}} \plus{} \frac {a_{k \minus{} 1}}{a_{k \minus{} 2}} \plus{} ... \plus{} \frac {a_2}{a_1})}^2 \geq 2k^2 \minus{} 4k \plus{} 2$

για καθε φυσικο αριθμο $ k$ με $ 2 \leq k \leq n$.

Ποτε ισχυει η ισοτητα;[/quote]

Αφου δεν βλεπω ακομα καμια λυση απο κανεναν βαζω μια βοηθια σε hidden

[hide]Χρησημοποιοντας την Ταυτοτητα του Euler στη δοθεισα ανισοτικη σχεση, προκειπτει η σχεση που επιτρεπει την αποδειξη της ανισοτητας κανοντας χρηση της ανισοτητας του Cauchy, αφου πρωτα χρησημοποιηθει η ανισοτητα: $ a^2 \plus{} b^2 \geq 2ab$[/hide]


I understand each team designs it's own t-shirt, yes?

Not always, some teams choose to, I think most teams do this.
I remember that Hawaii had really cool jackets last year (sports jackets)


Prove that if two coplanar circles intersect in two points that the line connecting these points bisects a common tangent of the two circles.

I just smash and bashed with Pythagorean Theorem.


Να λυθεί στους θετικούς πραγματικούς:

$ ab \plus{} bc \plus{} ca \plus{} 4 \equal{} 2(a \plus{} b \plus{} c) \plus{} abc$
$ a^3 \plus{} b^3 \plus{} c^3 \equal{} 3$

Αν δε λυθεί μέχρι αύριο θα βάλλω τη λύση σε hidden. Πάντως είναι δύσκολη. :wink:

Πολύ καλή άσκηση! Λοιπόν έχουμε και λέμε

Θέτουμε

$ S_1 \equal{} a \plus{} b \plus{} c$
$ S_2 \equal{} ab \plus{} bc \plus{} ac$
$ S_3 \equal{} abc$

Τότε επειδή $ (a \plus{} b \plus{} c)^2\geq 3(ab \plus{} bc \plus{} ac)$ (Ισχύει για οποιαδήποτε πραγματικά $ a,b,c$ αυτή η ανισότητα και ΟΧΙ μόνο για θετικά) άρα $ \boxed{S_1^2\geq 3S_2 \ \ (1)}$.

Επίσης η πρώτη σχέση γίνεται $ \boxed{S_2 \plus{} 4 \equal{} 2S_1 \plus{} S_3 \ \ (2)}$.

Επειδή
$ a^3 \plus{} b^3 \plus{} c^3 \minus{} 3abc \equal{} (a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} bc \minus{} ac) \equal{} (a \plus{} b \plus{} c)\left[(a \plus{} b \plus{} c)^2 \minus{} 3(ab \plus{} bc \plus{} ac)\right]$
άρα μετά από πράξεις (και αντικαθιστώντας το $ S_3$ από τη σχέση $ (2)$), καταλήγουμε στη σχέση

$ S_1^3 \minus{} 3S_1S_2 \plus{} 3S_2 \minus{} 6S_1 \plus{} 9 \equal{} 0$ απ' όπου εαν $ S_1\neq 1$ έχουμε $ S_2 \equal{} \frac {(S_1 \plus{} 3)(S_1^2 \minus{} 3S_1 \plus{} 3)}{3(S_1 \minus{} 1)}$

Όμως λόγω της σχέσης $ (1)$ παίρνουμε $ \frac {(S_1 \plus{} 3)(S_1^2 \minus{} 3S_1 \plus{} 3)}{3(S_1 \minus{} 1)}\leq\frac {S_1^2}{3} \Rightarrow \frac {(S_1 \minus{} 3)^2}{S_1 \minus{} 1}\leq 0$.

Εαν $ S_1\neq 3$ τότε πρέπει υποχρεωτικά $ S_1 < 1$. Θα δείξουμε ότι αυτό δεν γίνεται.

Από την ταυτότητα

$ 0\leq a^{4} \plus{} b^{4} \plus{} c^{4} \equal{} (a \plus{} b \plus{} c)(a^{3} \plus{} b^{3} \plus{} c^{3}) \minus{} (ab \plus{} bc \plus{} ac)(a^{2} \plus{} b^{2} \plus{} c^{2}) \plus{} abc(a \plus{} b \plus{} c)$ χρησιμοποιώντας την $ (2)$ για να απαλείψουμε το $ S_3$ και την βασική ταυτότητα $ a^2 \plus{} b^2 \plus{} c^2 \equal{} S_1^2 \minus{} 2S_2$ παίρνουμε ότι πρέπει να ισχύει πάντα ότι

$ 2S_2^2 \plus{} (S_1 \minus{} S_1^2)S_2 \plus{} 7S_1 \minus{} 2S_1^2\geq 0$ δηλαδή θα πρέπει η διακρίνουσα αυτού του τριωνύμου να είναι πάντα $ \leq 0$.

Δηλαδή (μετά από πράξεις) θα πρέπει $ S_1^4 \minus{} 2S_1^3 \plus{} 17S_1^2 \minus{} 56S_1\leq 0$ και σε συνδυασμό με το ότι $ S_1 < 1$ παίρνουμε $ \boxed{0\leq S_1 < 1 \ \ (3)}$ (Όταν $ S_1 < 0$ τότε $ S_1^4 \minus{} 2S_1^3 \plus{} 17S_1^2 \minus{} 56S_1 > 0$ που είναι άτοπο).

Δεν έχω ολοκληρώσει ακόμη την τελευταία περίπτωση $ (3)$ (διότι πρέπει να φύγω) η οποία είμαι σχεδόν σίγουρος ότι απορρίπτεται και έτσι τελικά

θα πάρουμε $ S_1 \equal{} 3$. Όμως τότε $ S_2 \equal{} 3$ και $ S_3 \equal{} 1$.

Τελικά λοιπόν τα $ a,b,c$ είναι λύσεις της εξίσωσης $ t^3 \minus{} S_1t^2 \plus{} S_2t \minus{} S_3 \equal{} 0$ δηλαδή της $ t^3 \minus{} 3t^2 \plus{} 3t \minus{} 1 \equal{} 0$ δηλαδή $ (t \minus{} 1)^3 \equal{} 0$ και έτσι η μοναδική λύση είναι η $ a \equal{} b \equal{} c \equal{} 1$.

Θα συνεχίσω την άλλη περίπτωση αύριο και μόλις βρω κάτι θα το στείλω

Αλέξανδρος

Αν βγει η τελευταία περίπτωση που λες τότε η εκφώνηση είναι σωστή. Εγώ όμως ξέχασα να πω στους θετικούς πραγματικούς, οπότε μην το ψάξεις διότι $ a \plus{} b \plus{} c < 1$ και $ a^3 \plus{} b^3 \plus{} c^3\equal{}3$ για θετικούς δε γίνεται.

Ζητώ συγγνώμη. :blush:

Kώστα πραγματικά ωραία άκσηση! Θέλω να μας πεις μετά και πώς την εμπνεύστηκες :idea:

Εγώ δεν βρήκα λύση αλλά έκανα ένα substitution που νομίζω βοηθάει πολύ :wink:

Θέτω $ x \equal{} a \minus{} 1, y \equal{} b \minus{} 1, z \equal{} c \minus{} 1$  :!:

Oι δοσμένες γίνονται:

$ xyz \plus{} x \plus{} y \plus{} z \equal{} 0$  :!:  kai

$ \sum{(x \plus{} 1)^3} \equal{} 3\iff \sum{x^3} \plus{} 3\sum{x^2} \plus{} 3\sum{x} \equal{} 0$

Mετά από εκεί δεν έχω προχωρήσει πολύ :( ...μια ιδέα π.χ. είναι να μετατρέψεις το $ \sum{x^3} \plus{} 3\sum{x} \equal{} \sum{x^3} \minus{} 3xyz$ kai na xrisimopoiiseis tin taytotita toy Euler.

Φυσικά η λύση του Αλέξανδρου πιθανότατα έχει αποτέλεσμα αλλά νομίζω ότι με τον τρόπο μου μπορούμε να αποφύγουμε αρκετές πράξεις :)

[color=red]Εdit: Αν η άσκηση αφορά μόνο τους θετικούς, μάλλον η αντικατάσταση που έκανα δεν είναι και τόσο πρακτική. αφού μετά τα $ x,y,z$ μπορεί να είναι και αρνητικά...[/color]

Η λύση του Αλέξανδρου είναι εκαρκέστατη και η προσέγγισή σου Δημήτρη ωραία είναι αλλά δεν ξέρω αν θα έχει αποτέλεσμα. Θέλει ψάξιμο.

Εγώ την εμπνεύστηκα από την ανισότητα 5.95 του βιβλίου του κύριου Στεργίου που είναι η εξής:

$ k,l,m > 0$

$ (\frac{2k}{k \plus{} l})^3$ και κυκλικά $ \geq 3$

Θέτουμε τώρα κάθε ένα από τα τρία κλάσματα $ a \equal{} \frac {2k}{k \plus{} l}$ και μετά από algebraic manipulations παίρνουμε την πρώτη σχέση του συστήματος οπότε είναι προφανές.

Αν χρειάζεται περισσότερη εξήγηση κανείς μη διστάσει και θα γράψω complete solution.
[color=red]
Moderator edit[ όταν θες να γράψεις >= γράψε \geq και όταν θες να γράψεις κλάσμα γράψε \frac{a}{b}][/color]

Η άσκηση είναι πραγματικά εντυπωσιακή! Μέχρι και το mathematica έκανε 2 δευτερόλεπτα για να τη λύσει με εντολή Reduce (λύνει τη μια εξίσωση και αντικαθιστά στην άλλη) ενώ η εντολή Solve (βρίσκει χρησιμοποιώντας όλες τις δυνατές αλγεβρικές μεθόδους τις λύσεις) κόλησε το μηχάνημά μου για σχεδόν ένα λεπτό!!!! Θέλω να επισημάνω ότι εκτός τη λύση (1,1,1) υπάρχουν άπειρες αρνητικές λύσεις, επομένως ορθώς η επιλογή για θετικούς. (Γιατί σωστή είναι η άσκηση όχι που έχει αρκετά στοιχεία για να γίνει ένας τρόπος λύσης αλλά που έχει αρκετά στοιχεία και για να δοθεί η λύση που ζητείται). Μπράβο και στη σκέψη.


inequalities
linear algebra
linear algebra unsolved

If $\ A\in M_n(\mathbb{R}),n\geq2$ and $\ Tr(AX)\leq TrA\cdot TrX$ for every $\ X\in M_n(\mathbb{R})$ prove that $\ A=O_n$.

the set of matrices $M$ such that $Tr(AM)=1$ is an affine hyperplane of $M_n(\mathbb{R})$, and so is the set of matrices $M$ such that $Tr(M)=0$.
If $A \neq \alpha Id$, then the intersection between these two hyperplanes will not be empty.
If $A = \alpha Id$, the conclusion is clear.

By changing X into -X one obtains: for all X tr(AX)=tr(X)tr(A).
The case  X=I implies tr(A)=0 and for all X tr(AX)=0.
(U,V)->tr(U.tV) is a scalar product; thus A=0.


Julian has weights of 1, 3 and 7 pounds. What is the smallest whole number of pounds he cannot weigh using these three weights and a balance scale?

I can't find anything easier than guess and check.

1=1.
3-1=2.
3=3.
3+1=4.
7-3+1=5.
7-1=6.
7=7.
7+1=8.
7-1+3=9.
7+3=10.
7+3+1=11.
Nothing makes $ \fbox{12}$.

It is interesting to note that any whole number amount less than the amount weighed from using all 3 weights is viable.

[quote=sushi1233]sorry for the bump,
but is there any formula for this? For two numbers, there is the Chicken McNugget Theorem, but what about 3 numbers?[/quote]

I don't think so...

[quote=sushi1233]sorry for the bump,
but is there any formula for this? For two numbers, there is the Chicken McNugget Theorem, but what about 3 numbers?[/quote]

The Chicken McNugget theorem is different for a couple of reasons. For one, the Chicken McNugget theorem would allow you to use many copies of each weight; in this version, you can only use one. Also, here negative numbers can be used.

[quote=bestzack66][quote=sushi1233]sorry for the bump,
but is there any formula for this? For two numbers, there is the Chicken McNugget Theorem, but what about 3 numbers?[/quote]

i think you just add the numbers up and add 1[/quote]

Not true. For example, if the weights were $1$, $2$, and $10$, there is no way to make $4$.

[quote=sushi1233]I don't know if this works, but maybe if all of the weights were odd, then we can just take the sum and add 1.[/quote]

[quote=sdpandit]

Not true. For example, if the weights were $1$, $2$, and $10$, there is no way to make $4$.[/quote]

[quote=DottedCaculator]it works whenever 3*weight n<=weight n+1 for all n[/quote]

Thanks


Consider the following methods of taxation:

1. Taxation of 1% of the wealth of every citizen every year.

2. Taxation of 50% of the wealth of every citizen upon their death.

Does one of the other generate more public revenue?

Does one or the other do more to benefit the economy?

Mathew, I thought you were against the flat tax?

I am not sure. First of all, I think it depends on how much an average individual lives. As far as what is better for the economy, I think that it depends on the exact numbers given like how much people die at a specific year and how old are they, etc etc. It also probably depends on what period of time this is being asked.

Btw, can this actually have a mathematical solution?

Sorry for posting twice. As far as revenue, I think once again it depends on the numbers (data). It depends on the wealth of an individual (aggregate of individuals) at different points in their life. Like for example, if you take one individual who lived 50 years. The first 45 years he was poor. Then the last 5 years he had huge incomes. In this case the 50% tax would probably much better for public revenue because the 50% of his income collected at the end is much greater then 1% every year. At least if the numbers are big enough. Maybe I am just being illogical here, but I think for given numbers, a model can be constructed, just like your post is called, optimization.

I am like other income structures better than a flat tax, but a subset of the services rendered by a government are those that protect wealth (as distinct from income/wages), so I personally believe that some part of an existing system of taxation should tax wealth to that degree.

I do believe there is a mathematical solution.  Think about how much revenue will be generated by each method.  Think about the way personal wealth grows.  Then add to the equation the amount of investment capital left in the system.  I think the answer has clear demonstration.  I am interested in seeing if anyone includes logic I didn't think about in my calculations.

You had a good[b] question[/b],Mr.Crawford. However,
you should have included[b] small fraction of additional[/b] taxes[b] related to affluent business enterprises whose  business
profits are very sustainning,economically. What do you think?

                                                      Respectfully,
                                                     A.Eddington[/b]

Number 1 is much better because of problems with defining ownership of wealth. It's not clear to me how to define wealth for Number 2 in the case of a married couple when one of the members dies.


How do I join? What are the qualifications/requirements? Who should I contact?

Check in the Maryland Community Forum

http://www.artofproblemsolving.com/Forum/index.php?f=240

I don't see any post regarding joining the team.

Make a topic asking, that's the most likely place you'll find out how.

too late for this year, but contact Eric Walstein at Montgomery Blair High School.  Also, I will add the entry to the how to join an arml wiki. ([url=http://www.artofproblemsolving.com/Wiki/index.php/How_to_join_an_ARML_team]link[/url])

I think I saw Calculus at ARML this year, so apparently he figured it out.


A boat has a green light ( with wavelength $\lambda = 500 nm$) on its mast. Wath wavelength would be measured and what colour would be observed for this light as seen by a diver submerged in water (index of refraction $n = 1.33$) by the side of the boat?

This seems quite simple unless I am falling for the trap.

Since the frquency of light depends only on that of the source, the frquency remains same even when light passes into water.
Now $ c_{water} =\ \frac{c_{vacuum}}{\mu} \ where\ \mu $ is the refractive index. Hence using $ c =\ \lambda \times f $ we get that
$ \lambda _{water} =\ \frac{\lambda}{\mu} $ .

I am not doing the calculation, but by these formulae you get the $ \lambda $ of light in water and can easily find from a table the corresponding colour.

finding the wavelenght n stuff is easy,but i have some doubts......
the light is green,the velenght will change,so will the colur of the light change?green will turn into some other colur?if it was whitelight,it would have changed because w light has seven colours,but green is ust green????? :(  :?:
is this a trick question?

Yes, the wavelength of the light IN WATER changes. But any observer, for example the diver, carries his own observing instruments (i.e. the eye) with its own medium (i.e. the yello - type thing inside the eye) that has its own and constant index of refraction and the color observed doesn't change.

Kubus,whats the yello thing?could u be more clear,please :lol:
so is ur answer that the light colour observed does not change?thats what i thought.i'm not sure 'bout it.
is my explaination right?


modular arithmetic
group theory
abstract algebra

If $(a,p)=1$ where $p$ is a prime number and

$n \equiv m \pmod{p-1}$ show that $a^{n}\equiv a^{m}\pmod{p}$

Use this result to find the last digit in the base 7 representation of $2^{1000000}$.
[Hint: $10^{n}\equiv 4 \pmod{6}$ for any positive integer $n$]

[b]Solution.[/b]
Let $G=(\mathbb{Z}/p\mathbb{Z})^{\times}$ be a multiplicative group, and let $H=\{1,a,a^{2},\hdots\}$ be a subgroup of $G$ generated by $a$. $\gcd (a,p)=1$ insures that $a\in G$.  The order of $a$, that is $\text{ord}_{p}a$, is equal to the order of $H$, so $a^{\text{ord}_{p}a}\equiv 1\pmod{p}$.

From $m\equiv n\pmod{\varphi (p)}$ we obtain $\varphi (p)|m-n$.  Since $|G|=\varphi (p)$, by Lagrange's Theorem, $\text{ord}_{p}a|\varphi (p)$, i.e. $\text{ord}_{p}a|m-n\implies m-n=k\cdot\text{ord}_{p}a$. Hence,
\[(a^{\text{ord}_{p}a})^{k}\equiv a^{m-n}\equiv 1\pmod p, \]
as required.$\blacksquare$


function
trigonometry
quadratics
algebra
complex analysis
complex analysis unsolved

1. What's the range of function cos(z)? Is that of whole complex plane? Is there an easy way to prove it?

2. Suppose{fn} n=1,...infinit, is a uniformly bounded sequence of functions of all analytic on the disc B(a,R). Suppose that for each z in B(a,R), the sequence converges pointwise. Show that the sequence {fn} converges UNIFORMLY on compact subsets of B(a,R).

1.  Yep.  For real $ x$, $ \cos x$ maps surjectively to $ [\minus{}1, 1]$ and $ \cos ix$ maps surjectively to $ \mathbb{R} \minus{} (\minus{}1, 1)$.

More on the first: $ \cos z\equal{}w\plus{}\frac1w$, where $ w\equal{}e^{ix}$ ranges over all nonzero complex numbers. $ w\plus{}\frac1w\equal{}c$ is a quadratic equation, which has two solutions unless it's a perfect square, and thus $ \cos z\equal{}c$ has two solutions per period except for $ c\equal{}\pm 1$, which have one solution per period each.


2. A circular table is pushed into the corner of a square room so that a point P on the edge of the table is 8" from one wall and 9" from the other wall as shown. Find the radius of the table in inches.

I don't know what to do..
[hide="Answer"]29 inches[/hide]

[quote="policecap"][hide](r-8)^2+(r-9)^2=r^2[/hide][/quote]
My thoughts:

[hide]
I think policecap's reasoning comes from the equation of a circle:
(x-h)^2+(y-k)^2=r^2 with (h,k) being the center.
In this case, h and k can both be r (if you move the circle to Quadrant I).
You can substitute (8,9) for (x,y) since (x,y) is a point on the circle.

You then get (8-r)^2+(9-r)^2=r^2, equivalent to his equation.
Expanding gets r^2-16r+64+r^2-18r+81=^2.
So r^2-34r+145=0
(r-29)(r-5)=0
[b]r=29[/b], since 5 is too small. [/hide]

theres a much simpler way to get that

btw where did that test come from and which year is it?

[quote="zolojetto"]isnt that the same thing that both posts before you said because using pythagorean theorum you get $ (x \minus{} 8)^2 \plus{} (x \minus{} 9)^2 \equal{} r^2$(exactly what policecap said) which gives you $ r \equal{} 29$.[/quote]

except distracted got that equation in a much more complicated way


Where can you get practice material for Mathcounts state and National?

Is anyone able to recommend the educaide program as being very helpful in preparing for the Mathcounts tests?  The website doesn't really focus on what you get for the $500.00.  That's a lot to spend if it isn't super helpful.
Do you actually get real past materials?
Also, correct me if I'm wrong, but The Mathcounts Store only has the 1999-2003 competition sets available for buying.
I'd love some earlier tests, but this is my school's second year, so we have almost no materials.
Any ideas under $500.00?!?

It is that much and maybe more, they sell it to schools, your school had it so your teacher could give it to you for free.

The Educaide softwae which is actually like $600 has no problems at all.  You then need to spend another $125 to $200 for the problems.  I bought it for my school.  It comes as a searchable computer database of every MATHCOUNTS handbook and competition problem written by MATHCOUNTS from like 1983 - 1998(i think).  I here they are finally working on updating the database with the more recent problems.  I love the program and it is a huge resource(over 10,000 problems) but it is very expensive!

Josh Frost
Coach
Coopertaive Middle School
New Hampshire

my school doens't have the educaide thing but it has a passkey system installed throughout the schools. it covers all the stuff from seventh grade to 12 th grade in reading writing (there are also other ares too like math).

you do't have to the ecucaide. the teacher at a nearby school had all the mathcounts stuff and i sort of did some of them he's won half the trips to national for the state of utah.

I briefly looked at the Educaide MATHCOUNTS module about 4 years ago.

There was a vast quantity of MATHCOUNTS material.  It appeared to be organized very well.

Among the online samples that were available, I thought there was a tendency to more typographical errors than in the original MATHCOUNTS material.

I have never actually used the product, and I do not know if the online samples were reflective of the actual product.  Would someone who has used the product care to comment?


limit
topology
real analysis
real analysis unsolved

in a metric space $ X$, let $ B_{r}(x)$ be a closed ball with center $ x\in X$ (and with radius $ r>0$)

prove that the metric space $ X$ is complete iff every sequence $ \{B_{r_n}(x_n)\}_n$, such that $ B_{r_{i\plus{}1}}(x_{i\plus{}1}) \subseteq B_{r_i}(x_i)$ and $ r_n \to 0$ (when $ n \to \infty$), has a non-empty intersection $ ( \text{i.e.}\; \bigcap_{n\equal{}1}^{\infty} B_{r_n}(x_n) \neq \emptyset\;\; )$

:D

If you had a sequence of balls satisfying that chain of containments, we'd have that $ x_m\in B_{r_n}(x_n)$ whenever $ m>n.$ That would make $ d(x_m,x_m)\le r_n,$ and since $ r_n\to0,$ that means that $ (x_n)$ is a Cauchy sequence.

If the metric space is complete, then there exists a point $ y$ such that $ \lim x_n\equal{}y.$ At this point it becomes important that you insisted on closed balls. For each $ n,$ we have $ x_m\in B_{r_n}(x_n)$ for all $ m>n,$ and that means that $ \lim x_m\in B_{r_n}(x_n).$ That means that $ y\in\bigcap_{n\equal{}1}^{\infty}B_{r_n}(x_n).$

The other way around: Suppose we have a Cauchy sequence $ (x_n).$ Choose a subsequence $ y_k\equal{}x_{n_k}$ such that $ d(y_k,y_{k\plus{}1})\le 2^{\minus{}k}.$ Let $ r_k\equal{}2^{\minus{}k\plus{}1}.$ I now claim that $ B_{r_{k\plus{}1}}(y_{k\plus{}1})\subseteq B_{r_k}(y_k).$ By hypothesis, this nested sequence of balls has a nonempty intersection. Let $ z$ be a member of that intersection. Then $ d(y_k,z)\le r_k.$ Thus $ \lim y_k\equal{}z.$ But if any subsequence of a Cauchy sequence converges, then the whole sequence converges and to the same limit.

It should be noticed that one consequence of the radii of the balls tending to zero, in both cases above the intersection of the nested sequence of balls actually consists of a single point.

I needed to pull out the idea of a subsequence whose consecutive differences are smaller than your favorite convergent series in order to prove the second part. It's too difficult to get a sequence of balls that actually nests if I don't do that.


There are lots of solutions, so what is the question?


Let $ x_1,x_2,\dots,x_n$ be positive real numbers. Let $ m\equal{}\min\{x_1,x_2,\dots,x_n\}$, $ M\equal{}\max\{x_1,x_2,\dots,x_n\}$, $ A\equal{}\frac{1}{n}(x_1\plus{}x_2\plus{}\dots\plus{}x_n)$, and $ G\equal{}\sqrt[n]{x_1x_2 \dots x_n}$. Prove that \[ A\minus{}G \ge \frac{1}{n}(\sqrt{M}\minus{}\sqrt{m})^2.\]

WLOG assume $ M\equal{}x_1,m\equal{}x_2$.
\[ A\minus{}G\ge\frac1n(\sqrt{x_1}\minus{}\sqrt{x_2})^2\\
\Leftrightarrow x_1\plus{}x_2\plus{}\ldots\plus{}x_n\minus{}n\sqrt[n]{x_1x_2\ldots x_n}\ge x_1\plus{}x_2\minus{}2\sqrt{x_1x_2}\\
\Leftrightarrow \sqrt{x_1x_2}\plus{}\sqrt{x_1x_2}\plus{}x_3\plus{}\ldots\plus{}x_n\ge n\sqrt[n]{x_1x_2\ldots x_n},
\]which is true by AM-GM. :D


conics
parabola
number theory proposed
number theory

Solve in integers x,y:

x^3+9xy+127=y^3.

[quote="bomb"]Solve in integers x,y:

x^3+9xy+127=y^3.[/quote]

if $f(x,y)=x^3+9xy+127-y^3$, we can check that $f(x,x+3)=100$. If $x>3$, then $f(x,x+4)=-3(x+7)(x-3)<0$, we can even see that for all $n>3$ $f(x,x+n+1)<f(x,x+n)$, because of the shape of the parabola $y=-3(n-3)(2x+n)$. So $x\in{{1,2,3}}$.
$f(1,y)$ and $f(2,y)$ don't have integer solutions. Solving $f(3,y)=0$  I get $y=7$. So the only solution is $(3,7)$.

note  Does anyone know its name if this method of finding solutions in integers by finding an upperbound for some of the unknown has a name because it seems to me that this is quite a usefull thing.


For $ a,b,c\in R$, and $ x,y\in R$, such that $ x^2\plus{}y^2\equal{}2$, prove the following inequality

$ \left|a(x^2\minus{}1)\plus{}bxy\plus{}c(y^2\minus{}1)\right|\leq \sqrt{(a\minus{}c)^2\plus{}b^2}$

this one can easily be solved by trigonometric substituting

Rewrite it as
\[ |(a\minus{}c)\frac{x^2\minus{}y^2}{2}\plus{}bxy| \le \sqrt{(a\minus{}c)^2\plus{}b^2}.\]
By Cauchy,
\[ |(a\minus{}c)\frac{x^2\minus{}y^2}{2}\plus{}bxy| \le \sqrt{\left((a\minus{}c)^2\plus{}b^2\right)\left(\left(\frac{x^2\minus{}y^2}{2} \right)^2\plus{}x^2y^2\right)}\equal{} \sqrt{(a\minus{}c)^2\plus{}b^2}.\]

Very nice solution, [b]FelixD[/b]! Good job! Thanks!

thanks for your gift, Marin.Bancos! Felix's solution is essentially the same as mine so needless to say, congrats on him!


modular arithmetic
algebra proposed
algebra

Compute
\[
\displaystyle\sum_{k=0}^n \left[\frac{2^k}{7}\right]
\]
[a] denotes the integer part of a real number a.

Since $[2^k/7]=2^k/7-\{2^k/7\}$  the solution is
\[
\displaystyle\sum_{k=0}^n [2^k/7]=\{\begin{array}{lll}
\displaystyle\frac{2^{n+1}-1}{7}-m-1/7\qquad \textrm{if} \qquad n=3m\\
\displaystyle\frac{2^{n+1}-1}{7}-m-3/7\qquad \textrm{if} \qquad n=3m+1\\
\displaystyle\frac{2^{n+1}-1}{7}-m-1\qquad \textrm{if} \qquad n=3m+2\\
\end{array}
\]


vector
probability
conditional probability
advanced fields
advanced fields unsolved

Suppose a random variable sequence in $\mathbb{R}^k$ is generated by:
$X_{n+1}=X_n+a_n \cdot d_n$. where $\{d_n\}$ are i.i.d unit-length random direction vectors whose direction is uniform distributed in $\mathbb{R}^k$. $\{a_n\}$ are postive scalar stepsizes satisfying $\sum a_n=\infty$ and $\sum {a_n}^2<\infty$.

Then what is the behavior of of $\{X_n\}$? does it converge in probability or with probability 1. Maybe the conclusion is dimension-dependent.

[quote="fedja"]$X_n$ converge with probability $1$. This is a classical theorem contained in most advanced probability books.  :)[/quote]

Not in my probability books :(. Could you give me a sketch of the proof or a referrence. I guess it invovles the convergence of martingale.

OK, here is how it is usually proved. You start with the following lemma
$P\{\sup_{1\leq n\leq N} |X_n|>t \}\leq 2 P\{|X_N|>t\}\leq 2t^{-2}E |X_N|^2=2t^{-2}\sum_{1\leq k\leq N}a_k^2$.
To prove the lemma, note that the event $\{\sup_{1\leq n\leq N} |X_n|>t \}$ is the union of the disjoint events $A_k=\{|X_k|>t\text{ and }|X_{j}|\leq t, j<k\}$. Now, since $X_N-X_k$ is independent of $A_k$ and symmetric, we see that the conditional probability $P\{|X_N|\geq |X_k|\,|\,A_k\}\geq \frac 12$, from where the conclusion follows immediately. Passing to the limit as $N\to\infty$, we obtain that  $P\{\sup_{n} |X_n|>t \}\leq 2t^{-2}\sum_{k}a_k^2$. Similarly, $P\{\sup_{n\geq m} |X_n-X_m|>t \}\leq 2t^{-2}\sum_{k\geq m}a_k^2$, from where it follows that the probability of the smaller event $E_{m,t}=\{\sup_{n',n''\geq m} |X_{n'}-X_{n''}|>2t\}$ is also not greater than $2t^{-2}\sum_{k\geq m}a_k^2$. Let now $E$ be the event that the sequence $X_n$ is not Cauchy. If $P\{E\}>0$, then for some $t>0$, we must have $P\{E_t\}>0$ where $E_t=\{\text{ for every }m\text{ there exist }n',n''\geq m\text{ such that }|X_{n'}-X_{n''}|>2t \}$. But $E_t\subset E_{m,t}$ for all $m$, so $P(E_t)\leq 2t^{-2}\sum_{k\geq m}a_k^2\to 0$ as $m\to\infty$. Hope this helps  :)


trigonometry
geometry
similar triangles
AMC

Hey guys, here is a triangle problem which I cannot solve:

In right triangle ABC, we have angle B as the right angle. Draw the altitude from vertex B to the base of the triange (that is BC). Label the point where the altitude meets AC as D. Draw a line from vertex A to BC and label the point where this line meets BC as E. We have EC = DC = AE = 1. Find the length of BC.

I tried this problem by extending another altitude from point E to AC, and then using similar triangles, however, there always seems to be a bit more that I need to consider. Perhaps trigonometry may be introduced since we know that triangle AEC is isoceles indeed....

Any help is appreciated.

Keta.

bugzpodder, can you explain how you deduced that Angle AEB is twice angle ACB?

Hi, do a little angle chasing.   its a direct consequence of AE=EC

bugzpodder wrote:[hide]
notice if we let angle ACB=t
then angle AEB=2t
by cosine law on AEC,

AC^2=2+4cos(2t)
BE=cos(2t)
so BC=cos(2t)+1
also BCD is similar to ABC, which means
BC/AC=CD/BC
BC^2=AC
hence (cos(2t)+1)^2=2+4cos(2t)
[/hide]
can you finish it off now?

Not quite.... the last equation should be

[hide] (cos(2t)+1)^4 = 2 + 4cos(2t) [/hide] since [hide] AC^2 = 2+4cos(2t) [/hide] ... sooo.... nasty quartic. yay.

But this should work:

[hide]

Let BC = x. Then let t = <BCD. So cos t = 1/x. Then cos 2t = 2/x^2 - 1.

However, then BE = cos 2t, so x = BC = cos 2t + 1 = 2/x^2

Therefore, x^3 = 2 and BC = x = cbrt(2).

 [/hide]

Question: You mentioned:

"Draw the altitude from vertex B to the base of the triange (that is BC)"

Now, from browsing these forums I have come to understand that my school's teaching of terminology can be off what everyone else does. So here is my terminology question: If B is a vertex, then how can an altitude be dropped from it to BC? Would it not be dropped to AC?
Thanks


There are $ n$ pawns on $ n$ distinct squares of a $ 19 \times 19$ chessboard. In each move, all the pawns are simultaneously moved to a neighboring square (horizontally or vertically) so that no two are moved onto the same square. No pawn can be moved along the same line in two successive moves. What largest number of pawns can a player place on the board (being able to arrange them freely) so as to be able to continue the game indefinitely?

We color the chessboard using the alternating black/white scheme such that the center square is white. Define the [i]deviation[/i] of a white square to be the smallest $ k$ such that the $ 2k\plus{}1 \times 2k\plus{}1$ square centered at the center of the chessboard contains that white square; and let the deviation of a black square be undefined.

We note that each turn, a pawn moves from a square to a square of a different color, and that if a pawn is on a white square, two turns later it will move to a white square with a deviation that has a different parity. Hence, the parity of the deviation is periodic $ \mod 4$.

There are $ \left(\frac{19 \minus{} 1}{2}\right)^2$ white squares with an even deviation. Thus, there can be at most $ 4 \cdot \left(\frac{19 \minus{} 1}{2}\right)^2 \equal{} \boxed{18^2}$ pawns total.

To show that such a configuration exists, note (for example) that a $ 2\times 2$ square filled with $ 4$ pawns works (through a clockwise cycle). Then place our pawns on $ 9 \cdot 9$ of the $ 2 \times 2$ squares on the $ 19 \times 19$ chessboard.


inequalities
pigeonhole principle
inequalities unsolved

If the numbers a, b, c are all nonnegative, then this follows from the inequality given in http://www.mathlinks.ro/Forum/viewtopic.php?t=19666 when you set k = 1.

In the case of arbitrary real a, b, c, you can apply the inequality to the nonnegative numbers |a|, |b|, |c| to get

$\left(\left|a\right|^2+2\right)\left(\left|b\right|^2+2\right)\left(\left|c\right|^2+2\right) \geq 9\left(\left|a\right|\left|b\right|+\left|b\right|\left|c\right|+\left|c\right|\left|a\right|\right)$.

Since $\left|x\right|^2 = x^2$ for any real x, this simplifies to

$\left(a^2+2\right)\left(b^2+2\right)\left(c^2+2\right) \geq 9\left(\left|a\right|\left|b\right|+\left|b\right|\left|c\right|+\left|c\right|\left|a\right|\right)$.

And since $\left|x\right|\geq x$ for any real x, this yields

$\left(a^2+2\right)\left(b^2+2\right)\left(c^2+2\right) \geq 9\left(\left|a\right|\left|b\right|+\left|b\right|\left|c\right|+\left|c\right|\left|a\right|\right) \geq 9\left(ab+bc+ca\right)$.

Thus, your inequality is proven for arbitrary reals a, b, c.

  Darij

I think it was Vas_1 or Vasc_3 delightful inequality before crash :(
I solved them both, but unfortunately we lost these topics.

Anyway, it is quite easy ;)  :P  :P  :P  :P
(my favorite argument)

It is Schur and a bunch of tiny details (all sums are symmetric here)

The expansion of the $LHS$ is $x^2y^2z^2+2\sum x^2y^2+4\sum x^2+8$

$3\sum x^2\geq 3\sum xy$ by rearrangement

$2\sum x^2y^2+6\geq 4\sum xy$ by AM-GM

Finally, $x^2y^2z^2+\sum x^2+2\geq \sum x^2+3(xyz)^{\frac {2}{3}}\geq \sum x^{\frac {4}{3}}y^{\frac {2}{3}}\geq 2\sum xy$ by AM-GM, Schur, and Muirhead in succession.

Therefore, $LHS\geq RHS$

Maybe someone smarter could tell me if this solution is OK ?

3RHS = (x^2 + y ^2 + 1 +x^2 + y^2 + x^2y^2 + 3)(1 + 1 + z^2 + z^2 + z^2 + 1 + 3) >= (x + y + z + xy + xz + yz + 3)^2 (using Cauchy-Schwartz)
let xy + xz + yz = a^2
then x + y + z >= a*sqrt(3)
Now (x + y + z + xy + xz + yz + 3)^2 >= (a^2 + a*sqrt(3) + 3)^2
by AM >= GM -> (a^2 + a*sqrt(3) + 3)^2 >= [3a*sqrt(3)]^2 = 27a^2 = 3LHS
but now 3RHS >= 3LHS, and we are done.

@[b]mathboy_cntHDVN[/b]: You [b]spam[/b] to much. Please see the day when this topic was posted.

Put $ t \equal{} \frac {b \plus{} c}{2}$
By AM-GM we have:
$ (b^2 \plus{} 2)(c^2 \plus{} 2) \geq 3 \plus{} 2bc \plus{} 2(b^2 \plus{} c^2) \geq 3 \plus{} 6t^2$ and $ t^2 \geq bc$
We need prove that:
$ (a^2 \plus{} 2)(3 \plus{} 6t^2) \geq 9(2at \plus{} t^2)$
<=>$ (a \minus{} t)^2 \plus{} 2(at \minus{} 1)^2 \geq 0$
This inequality is true
Equality occurs when  $ a \equal{} b \equal{} c \equal{} 1$

[quote="pigfly"]Let $ a,b,c\in\mathbb{R}.$ Prove that $ \left(a^2 \plus{} 2\right)\left(b^2 \plus{} 2\right)\left(c^2 \plus{} 2\right)\geq 9(bc \plus{} ca \plus{} ab).$[/quote]$ \prod{\left(a^2 \plus{} 2\right)} \minus{} 9\sum{bc} \equal{} \frac {8}{9}(bc \plus{} ca \plus{} ab \minus{} 3)^2 \plus{} \frac {35}{18}\sum{(b \minus{} c)^2}$

$ \plus{} \frac {1}{27}\sum{\left[a \plus{} b \plus{} c \minus{} 3abc \plus{} \sqrt {5}(ca \plus{} ab \minus{} 2bc)\right]^2}\geq0.$

This problem is introduced as Example 2.1.1 on pp. 34-35 in new book :

Ji Chen, Chao-Cheng Ji, [url=http://www.jzbook.net.cn/list/bookdetail.asp?prodid=O1568]Algebraic Inequalities (Mathematical Olympiad Proposition people Lecture)[/url], Shanghai Scientific & Technological Educational Publishing House, 2009, 225p. (ISBN 978-7-5428-4848-2/O·613)

See also : http://www.mathlinks.ro/viewtopic.php?t=88927

[quote="pigfly"]Let $a,b,c\in\mathbb{R}$.Prove that:
\[(a^2+2)(b^2+2)(c^2+2)\geq 9(ab+bc+ca)\]
Sorry if it was posted before! ;)[/quote]
By Pigeonhole, two of $a,b,c$ are both greater than $1$ or both less than $1$. Let those be $a,b$.
\[(a^2+2)(b^2+2)-3(a^2+b^2+1) = 1 - a^2 - b^2 +a^2b^2 = (1-a^2)(1-b^2) \ge 0\]
Therefore, $(a^2+2)(b^2+2) \ge 3(a^2+b^2+1)$.
But, we have \[9(ab+bc+ca) \le 3(a+b+c)^2 \le 3(a^2+b^2+1)(1+1+c^2) \le (a^2+2)(b^2+2)(c^2+2)\]
We are done.

[quote="pigfly"]Let $a,b,c\in\mathbb{R}$.Prove that:
\[(a^2+2)(b^2+2)(c^2+2)\geq 9(ab+bc+ca)\]
Sorry if it was posted before! ;)[/quote]
You can see http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=453747


Of all oredered triples of positive integers (x,y,z) that satidfy 2^3x+2^4y=2^5z, find the ordered triple with the smallest value for z.

[hide="Solution"]Two powers of two can only sum to a power of two if and only if they are equal. This is obvious from analyzing $ 2^a \plus{} 2^b$, $ 2^a \plus{} 2^b \equal{} 2^a(1 \plus{} 2^{b\minus{}a})$, which is only a power of two if $ 2^{b\minus{}a}$ is one less than a power of two, which happens only when $ b \equal{} a$ because the only consecutive powers of two are $ 1$ and $ 2$.

Therefore, $ 3x \equal{} 4y$, and $ 5z \equiv 1$ modulo $ 3$ and $ 4$. This is equivalent to $ 5z \equiv 1$ modulo $ 12$ by the Chinese Remainder Theorem, which is equivalent to $ z \equiv 5$ modulo $ 12$. Therefore, the smallest possible value of $ z$ that is valid is $ \boxed{5}$. Let $ x \equal{} 8$ and $ y\equal{}6$, and such a value of $ z$ is confirmed.[/hide]


induction
Pascal\u0027s Triangle
combinatorics proposed
combinatorics

Let $n \geq 2$ $n \in \mathbb N$ and $M=\{ 1,2,3,..,n \}$
$x_{k}=\frac{1}{n+1}\cdot \sum^{|A|=k}_{A\subset M}(min A+max A)$ ,$k\in \{1,2,..,n-1 \}$

where $|A|$ is the cardinal of $A$
Prove that $x_{k}\in \mathbb N$ and not all of $x_{k}$ are divisible by 4.

How many sets $A$ with $k$ elements such that a fixed number $t = \min A$ are there? We first choose $t$ and then the other $k-1$ elements from $\left\{ t+1, \ldots, n \right\}$.
How about $t = \max A$? Well, we first choose $t$ and then the other $k-1$ elements from $\left\{ 1, \ldots, t-1 \right\}$.
Hence, $x_{k}= \sum_{i=1}^{n}i \cdot \binom{n-i}{k-1}+\sum_{j=1}^{n}j \cdot \binom{j-1}{k-1}$.
Now make the "substitution" $j-1 = n-i$ in the second sum:
\begin{eqnarray*}(n+1)x_{k}&=& \sum_{i=1}^{n}i \cdot \binom{n-i}{k-1}+\sum_{i=1}^{n}(n-i+1) \cdot \binom{n-i}{k-1}\\ \ &=& \sum_{i=1}^{n}(n+1) \cdot \binom{n-i}{k-1}\\ \ &=& (n+1) \sum_{i=1}^{n}\binom{n-i}{k-1},\end{eqnarray*}
so $x_{k}= \sum_{i=1}^{n}\binom{n-i}{k-1}= \sum_{j=1}^{n}\binom{j-1}{k-1}= \binom{n}{k}$ :what?: (Fix $k$ and use induction over $n$ to prove this.)

After I return from school, I'll try to solve the second part also.

For any number $s$ is a minimum element of $\binom{n-s}{k-1}$ subsets and a maximum element of $\binom{s-1}{k-1}$ subsets.So
$\sum^{|A|=k}_{A\subset M}min A=\binom{n-1}{k-1}\cdot1+\binom{n-2}{k-1}\cdot2+...+\binom{k-1}{k-1}\cdot (n-k+1)=\binom{n+1}{k+1}$
$\sum^{|A|=k}_{A\subset M}max A=\binom{k-1}{k-1}\cdot k+\binom{k}{k-1}\cdot(k+1)+...+\binom{n-1}{k-1}\cdot n=k\cdot\binom{n+1}{k+1}$
$\Longrightarrow x_{k}=\frac{k+1}{n+1}\binom{n+1}{k+1}=\binom{n}{k}\in \mathbb N$
For odd $n$ taking $k=1$ gettin $x_{k}$ odd.
For even $n$ let $n=2^{\alpha}p$ and $k=2^{\alpha-1}$ then $x_{k}=\binom{2^{\alpha}p}{2^{\alpha-1}}\Longrightarrow x_{k}$ isn't divisible by 4.

Now, the second part:

Let's think about [url=http://en.wikipedia.org/wiki/Pascal%27s_triangle]Pascal's Triangle[/url]. Suppose that on $n$-th line there are elements not divisible by $4$, besides the 1's (the set of such $n$ is clearly non-empty), but on the $n+1$-th line such elements do not exist. Knowing that each element on $n+1$-th line is the sum the two above it, we deduce that the $n$-th line must look like this (modulo 4): $1, 3, 1, 3, 1, \ldots \ \ \left( \star \right)$ Since it must end in a $1$, $n$ is even, so $n=2k$. But $\binom{2k}{k}$ is even, because when we partion $\left\{ 1, 2, \ldots, 2k \right\}$ into two sets of equal cardinality, we find two combinations of $k$ elements. Using $\left( \star \right)$ we get that the same number must be odd, contradiction.


hey mayb this is an easy problem,but it sticking like a thorn in my flesh (if the moderators feel it's easy, move it to the beginners section)
in a triangle ABC, D,E,F  are points on the sides BC,CA,AB. also A,B,C are points on sides YZ,ZX,XY of triangle XYZ for which FE paralle to YZ, FD parallel to ZX, DE parallel to XY.
Prove that $\left[ABC\right]^{2}=\left[DEF\right]\cdot\left[XYZ\right]$.

[LMN] denotes area of triangle LMN.

Triangles DEF and XYZ have parallel sides so they are homothetic, so there exists $\{ O\}=ZF\cap XD\cap YE$.
We have $\displaystyle \frac{OF}{OZ}=\frac{OD}{OX}=\frac{OE}{OY}=\frac{1}k$.

$FE\parallel ZY\Longrightarrow \left[ FAE\right] =\left[ YFE\right] \Longrightarrow \left[ AFOE\right] =\left[ FOY\right] $
We have $\displaystyle \frac{\left[ FOY\right] }{\left[ FOE\right] }=\frac{OY}{OE}=k$.
So $\left[ AFOE\right] =k\left[ FOE\right] $.

By summing this and the others similar to this we get:
$\left[ ABC\right] =k \left[ D E F\right] $
So use $\displaystyle \frac{\left[ XYZ\right] }{\left[ D E F\right] }=k^{2}$ to get
$\displaystyle \left[ ABC\right] ^{2}=k^{2}\left[ D E F\right] ^{2}=\frac{\left[ XYZ\right] }{\left[ D E F\right] }\cdot \left[ D E F\right] ^{2}=\left[ XYZ\right] \cdot \left[ D E F\right] $.


This statement is true.

If you aren't sure whether a statement is true or not, test it out by plugging in some numbers and that'll give you an idea.

[quote="CruS"]sqrt(-x*-x) = -sqrt(x)*-sqrt(x) ?

if not, why?

thx[/quote]

I don't think it works for negative numbers:

$x=-2$

$\sqrt{-(-2)*-(-2)}=\sqrt{4}=\pm2$

$-\sqrt{-2}*-\sqrt{-2}$

$-2$

Hmmm.... :huh:

Oops. Drunken Math is correct. I wasn't thinking properly.  :blush:

[quote="CruS"]sqrt(-x*-x) = -sqrt(x)*-sqrt(x) ?

if not, why?

thx[/quote]

Did you already post that in another topic of you? I belive it is 1=-1 right?


Prove that, if $(n-1)|k$, then $(n-1)^{2}|(n^{k}-1)$, $n,k$ are natural numbers.

n^k -1 = (n-1)(a)

where a is the other polynomial in this well-known factorization.

So, we divide out and get

n-1, a

a= n^(k-1) + n^(k-2) + ..+1

where to go from here?

Continuing neelnal's suggestion:
We can employ the Polynomial Remainder Theorem.
We wish to prove that $(n-1)^{2}|n^{k}-1$, which can also be expressed by \[n-1|\frac{n^{k}-1}{n-1}\]  \[n-1|n^{k-1}+n^{k-2}+...+n^{1}+n^{0}\] Now define $f(n) = n^{k-1}+n^{k-2}+...+n^{1}+n^{0}$. Using the remainder theorem, the remainder of $f(n)$ divided by $n-1$ is the same as $f(1)$. \[f(1) = 1^{k-1}+1^{k-2}+...+1^{1}+1^{0}= k\] Thus the remainder is $k$ which is also divisible by $n-1$, causing the remainder to be effectively $0$.   $\mathbb{Q.E.D}\blacksquare$


linear algebra
matrix
combinatorics unsolved
combinatorics

A square matrix $2005\times 2005$ is filled with 1 and 0. It is known that for each unit the quantity of units in the same column OR in the same row is at most 40 percents. What is maximal number of units in the matrix?

i don't think i understood the problem-is it that every row and every column has atmost $802$ ones in it?($40$% of $2005$)

No, it is an absolutely wrong understanding.
If $a_{ij}=1$, then the condition states that either $i$-th row OR $j$-th column contains at most 40% of '1'.

Something like that.
Unfortunately, I even didn't find this answer in 1.5 hours.

I dont know if it is going somewhere, let $f_1,f_2...,f_{2005}$ be the sum in the rows, and $c_1,c_2...,c_{2005}$ be the sum in the columns.

Now supposse $c_1,c_2...,c_k>802>=c_{k+1}...,c_{2005}$, $f_1,f_2...,f_k>802>=f_{k+1}...,f_{2005}$

then for $c_i$, $i=1,2,3...k$, there are $c_i$ rows with less than $802$ elements, and each of this rows kills at most $f_t$ intersections.

From other point of view, the $c_1+c_2+...c_k$ squares with a $1$ in the columns $1,2,3...,k$, must be in a row, with less than $802$ $1$'s, and  row $t$ with at most that number of $1$'s contains at most $f_t$ such squares, so we get that: $c_1+c_2+...c_k<= f_{k+1}+...f_{2005}$ and analogously that: $f_1+f_2...+f_k<=c_{k+1}+...c_{2005}$ So it is easy to see the number of squares is at most:

$f_{k+1}+...f_{2005}+c_{k+1}+...c_{2005}$

 ;)  ;)

I dont know what else...because i dont even know if i am on a good way... :(


geometry
probability
function
expected value
probability and stats

I thought about putting it on the Round Table but this forum looks more suitable.

I'm taking statistics class through BYU Independent Study now. My school, which is in very rural area, paid around 500 dollars, including the online course and the book, for me to take this class. I started to do the class little bit before the midpoint of the school year and well, I only submitted one assignment so far (from every two chapters there is an assignment), which I got 70%. Frankly, I do not get what I'm doing in this chapter either (#3) and that it seems like I'm wasting my time.

I'm asking for sincere help on what I should do. I have up to this year's September to finish the course but I strongly doubt that I will. And I honestly feel that it's a waste for me when I'll barely pass or if my college doesn't accept it. I don't know.. It may be simple like "Just keep doing the course and post any question." But it's not. I'm having so much difficulty with the course and along with everything that I"m doing as a senior year and keeping my rank, I just don't have time to do it.

What should I do? Like... What is a solution to this case if you were me?

[quote="fedja"]Did you consider getting a tutor? If you find somebody really good, it may help enormously. Also, despite it'll cost some money, you'll know that you at least made every possible effort to succeed, so you'll not feel so bad about 500 dollars paid by your school in the end even if you fail (I don't know if it is a moral issue for you but it would be for me). As to wasting time, I'm not completely sure; it depends on whether you want to learn something similar to this course in the future. If yes, I would try to go up to the end, so the next time you take a similar course it wouldn't be scary at all. As to having no extra time, unfortunately, there is no remedy now; that problem had to be considered when you planned your activities in the beginning of the term :(. What are you doing now in that course, by the way, that seems so difficult?[/quote]

Tutor is out of the question because there is no one at this level to help me in regular basis around my area.

I do regret that I have a busy schedule but there really isn't anything to do it. I'll try to go to the end I suppose.

I'm learning about PDF (probability distribution function) and expected value stuff. I kind of get it and I kind of don't... well, I can't really explain it because I've always either understood the problem or not understood at all in other classes (latter came like once though...)

Ah, I see now where [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=190651]this post[/url] comes from. We can try to discuss it in that thread if you wish but can you be a bit more specific and bring up one topic a time? Both concepts are not that complicated (I was afraid you were doing something much more advanced, which I either never knew, or blissfully forgot long ago). I'll try to answer that post of yours starting with probability distribution function but it would help a lot if you could somehow indicate where your difficulty is. So, I'll start the explanation (perhaps, tomorrow) but at some point I'll stop and ask you a question :)


Euler
function
algorithm
number theory unsolved
number theory

this is new problem

find all $a$ in set $\mathbb{N}$ such that:
$2^{a+1}+1$ divisble a

I think this problem very hard

Let $p$ is a prime such that $p \mid 2^{p+1}+1$. Then $2^{p+1}+2^{p-1}\equiv 0 \bmod p$. $2^{p-1}\cdot 5 \equiv 0 \bmod p$. Hence $p = 5$.

i don't known why a must be a power of 5

Let $\phi(a)$ - Euler function then all solutions satisfies $2^{a+1-\phi(a)}+1\equiv 0 \bmod a$.
Hence if $a$ is prime then $5\equiv 0 \bmod a$, then $a=5$. But it is correct if $a$ is prime only.
These reasoning needs to be repeated for $2^{a+1-\phi (a)}+1\equiv 0 \bmod a$ and so on -contradiction .

juna wrote wrong

$2^{a+1-\phi(a)}+1$ divisble $a$
this is wrong...........

Let $2^{a+1}+1\equiv 0\bmod a$. If $a$  - odd then $2^{\phi(a)}\equiv 1\bmod a$. Hence $2^{a+1}+2^{\phi(a)}\equiv 0 \bmod a$, hence $2^{\phi(a)}(2^{a+1-\phi(a)}+1)\equiv 0 \bmod a$, hence $2^{a+1-\phi(a)}+1\equiv 0 \bmod a$ and so on - contradiction.

Sorry. I can't understand this:

    why can't  [img]http://www.mathlinks.ro/Forum/latexrender/pictures/4de28e8b6743b1b38f589e6a6b2fefb4.gif[/img] for natural numbers a.

Can someone tell me?

juna your solution is wrong - I don't have clue what do you mean by "and so on" and where do you see the contradiction.

Andreas solution works for primes only...

If $2^{a+1-\phi(a)}+1 \equiv 0 \bmod a$ then $2^{a+1-\phi(a)}+2^{\phi(a)}\equiv 0 \bmod a$ hence $2^{\phi(a)}(2^{a+1-2\phi(a)}+1) \equiv 0 \bmod a$ hence $2^{a+1-2\phi(a)}+1 \equiv 0 \bmod a$ and so on. $\forall n\in N (2^{a+1-n\phi(a)}+1) \equiv 0 \bmod a$ - contradiction. It is a method of infinite descent

INDEED,I SITLL don't known if n is a composite number then............

Juna your solution is wrong. For example if [img]http://www.mathlinks.ro/Forum/latexrender/pictures/b704d1d942412e3b2eec2979d45aa6ca.gif[/img] divides a+1 then there is no contradiction.

Where is this problem from? Because I'm not sure if it can be solved: for example it's easy to note that $a=377$ works...

Estimation for $n$
$\frac{a+1}{a-1}\leqslant n\leqslant \frac{a+1}{\phi(a)}-\frac{ln(a-1)}{\phi(a)ln(2)}$

[quote="TomciO"]Where is this problem from? Because I'm not sure if it can be solved: for example it's easy to note that $n=377$ works...[/quote]
It is easy to show, that a=337 don't work.
Let \[a=\prod p_{i}^{k_{i}}\] , then a is odd and \[k_{i}\le ord_{p_{i}}(2^{p_{i}}-1)\] . Because $2^{a+1}+1=(2^{(a+1)/2})^{2}+`1^{2}$ we have $p_{i}=1(mod 4)$. Let $T_{i}$ are minimal numbers, suth that $2^{T_{i}}=1(mod \ p_{i})$. It give $ord_{2}(T_{i})=k=ord_{2}(a+1)+1=2$, because $a=1(mod \ 4)\Longrightarrow ord_{2}(a+1)+1=2$. For example $ord(T_{5})=2=ord_{2}(T_{13})=ord_{2}(T_{29})=...$
For example works $a=5*13.$

Yeah, $337$ doesn't work but $377$ does...

If find $a=p_{1}p_{2}$, $p_{1},p_{2}$ - prime then $2^{p_{1}p_{2}+1}+1\equiv 0 \bmod{p_{1}p_{2}}\Longrightarrow 2^{p_{1}}2^{p_{2}}+1\equiv 0 \bmod{p_{1}p_{2}}$.
Using $\forall p\in P,p\neq 2(2^{p}\equiv 2 \bmod p)$ it is tested solutions - 65=5*13, 377=13*29. It is correct and I made the mistake. Sorry.

Can be show that this problem had infinetely many solutions. But $a=5*29$ does not work.

[quote="Rust"][quote="metehan"]YES You Saying True Tomci0. Maybe we can solve this problem for square free numbers n. I think it a is good start for this problem.[/quote]
Let $a=5*109*1093^{2}$. We have $T_{5}=4,T_{109}=36,T_{1093^{2}}=364$, therefore $a|2^{a+1}+1$ and a is not square free.[/quote]
  There is I said if we solve it for square numbers then maybe we will have good start for the problem.So I didn't say there is no solutions n for n is not square free. I know there is solutions for some not square free numbers n. :P

It was interesting yet knew already of retrieved solutions.
The algorithm the generating solutions and problem on perpetuity of the solutions is now interesting.


vector
geometry
parallelogram
real analysis
real analysis unsolved

Let X be vector space and $ \|\cdot\|$ be norm on X. Prove that if parallelogram law holds for all x,y in X, i.e. $ \|x\plus{}y\|^2 \minus{} \|x\minus{}y\|^2 \equal{} 2(\|x\|^2\plus{}\|y\|^2)$, prove there exist inner product $ <\cdot,\cdot>$ on X such that, for all x in X, $ \|x\|\equal{}\sqrt{<x,x>}$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=224293]Link[/url] to another thread, which contains another useful link.

The link is forbidden...is there any other information on this?

Reconstructing some of what was on that, as a series of exercises:

To keep it a little simpler, I'll assume that we're talking about a real vector space.

Define $ \langle x,y\rangle \equal{} \frac14\left(\|x \plus{} y\|^2 \minus{} \|x \minus{} y\|^2\right).$

1. Show that $ \langle x,y\rangle\equal{}\langle y,x\rangle.$

2. Show that $ \langle x,x\rangle\equal{}\|x\|^2$ and thus $ \langle x,x\rangle >0$ for $ x\ne 0.$

3. Use the parallelogram identity to prove that $ \frac{\|a\|^2\plus{}\|b\|^2}2\equal{}\left\|\frac{a\plus{}b}2\right\|^2\plus{}\left\|\frac{a\minus{}b}2\right\|^2.$

4. Prove that $ \langle x,0\rangle\equal{}0$ for all $ x.$

5. Prove that $ \langle x,y\rangle\equal{}\minus{}\langle x,\minus{}y\rangle$ for any $ x,y.$

6. Use the parallelogram identity and the result of #3 to prove
\[ \langle x\plus{}y,z\rangle\plus{}\langle x\minus{}y,z\rangle \equal{} 2\langle x,z\rangle.\]
Conclude that $ \langle a,z\rangle\plus{}\langle b,z\rangle\equal{}2\langle(a\plus{}b)/2,z\rangle$ and in particular that $ \langle 2x,z\rangle\equal{}2\langle x,z\rangle$ for all $ x, z.$

7. Using #6, show that $ \langle x\plus{}y,z\rangle\equal{}\langle x,z\rangle\plus{}\langle y,z\rangle.$

8. Show that $ \langle mx,z\rangle\equal{}m\langle x,z\rangle$ for all $ x,y$ and all $ m\in \mathbb{N}.$ (This should be an induction.)

9. Show that $ \langle rx,z\rangle\equal{}r\langle x,z\rangle$ for all $ x,y$ and all $ r\in \mathbb{Q}.$

10. Use the parallelogram identity to prove that for any $ x,z,$ $ a\mapsto \langle ax,z\rangle.$ Conclude that $ \langle ax,z\rangle\equal{}a\langle x,z\rangle$ for all $ x,y$ and all $ a\in \mathbb{R}.$

wow thx. actually i found the original paper as well. it was on pg 719-723 of ann. of math vol 36.


Se pare ca problema este neinteresanta. :(

Sincer, da. Oau, ai renuntat la  ;)  intr-un post?  :P

Sa stii ca nu numai la un post,la mai multe.Doar nu-s chiar asa de tampit  incat sa folosesc tot timpul  ;)  cand nu are rost

[quote="cezar lupu"]Iata ca gasi, uitandu-ma prin gazeta matematica, seria pentru tineret, o problema extrem de simpla,dar draguta dupa umila mea parere. Iata enuntul ei:

   Sa se decida care din urmatoarele numere este mai mare?

        $a=7^{\frac{1}{2}} +7^{\frac{1}{7}} $ si  $b=7^{\frac{1}{3}}+7^{\frac{1}{6}}$ ;)[/quote]

sper ca nu am gresit:

Lema 1: 0 < ln 7 < 2
Dem: $ e^0 =1 < 7 < 7.3441 = 2.71 ^ 2 < e^2 $

Lema 2: $ h(u)=u^2 7^u $ este crescatoare pe (0,1)
Dem: $h' (u)=u 7^u [ 2- (ln 7) u ] > 0$

Lema 3: $gg(x) = (-ln 7) \frac { 7 ^ \frac 1 x}  {x^2}$ este crescatoare pe $(0, \inf)$
Dem: $gg(x)=(-ln 7) h( \frac 1 u)$, cum h crescatoare si gg crescatoare

Lema 4: daca $g(x) = 7 ^ \frac 1 x $ si $f(x) = g(x) - g (x-1)$, atunci f crecatoare
Dem: g' = gg => g' crescatoare => f' pozitiva => f crescatoare

Corolar: a>b
Demonstratie: a-b = f(7) - f(3) > 0.

E mai simplu: daca $f$ e convexa (si $f(x)=7^{\frac1x}$ este pe $(0,+\infty )$), atunci pentru $a<b<c<d$ astfel ca $a+d=b+c$ avem $f(a)+f(d)>f(b)+f(c)$.

[u]Generalizare.[/u] Notam: $A.a.s.B\Longleftrightarrow sgnA=sgnB\Longleftrightarrow AB>0\ \vee \ A=B=0$
(A,B au [b][u]a[/u][/b]celasi [b][u]s[/u][/b]emn). Atunci $0<a<b<c<d,\ b+c<a+d;\ 0<X\Longrightarrow$

$\left[ \left( \sqrt [b] X +\sqrt [c] X\right) -\left( \sqrt [a] X +\sqrt [d] X\right) \right].a.s.(1-X).$

[u]Demonstratie[/u] (Ideea apartine Domnului [i]Marius Damian[/i]).

$\underline {Caz\ 1: \ 0<X<1.}\ \ \ 0<X^c <X^a,\ 0<X^{d-c}-1<X^{b-a}-1$ (produs)$\Longrightarrow$

$0<X^c\left( X^{d-c}-1\right) <X^a\left( X^{b-a}-1\right)\Longrightarrow X^d-X^c<X^b-X^a$ $\Longrightarrow X^a+X^d<X^b+X^c;$

$\underline {Caz\ 2.\ 1<X.}\ \ \ 0<X^a<X^c,\ 0<X^{b-a}-1<X^{d-c}-1\Longrightarrow$ (produs)

$0<X^b-X^a<X^d-X^c\Longrightarrow X^b+X^c<X^a+X^d.$

[u]Observatie.[/u] In ipoteza $0<a<b<c<d,\ b+c<a+d$ ecuatia
$x^a+x^d=x^b+x^c,\ x>0$ are o singura solutie, $x=1$.


[quote="K Sengupta"]Consider three positive integers x, y and z in Arithmetic Progression with x < y < z.
Determine all possible solutions of this equation:
x^3 + y^3 = z^3 - 2[/quote]
Y = X + A and Z = X + 2A for some positive integer A.
Therefore, the equation translates to
X^3 + (X+A)^3 = (X+2A)^3 - 2

X^3 + X^3 + 3AX^2 + 3A^2*X + A^3 = X^3 + 6AX^2 + 12A^2*X + 8A^3 - 2

X^3 - 3AX^2 - 9A^2*X - 7A^3 + 2 = 0

Now I'm stuck.  I don't know what positive integers A might make that equation factorable.

This is definitely NOT a puzzle.  It belongs under one of the "Algebra" sections.


geometry
circumcircle
symmetry
geometry unsolved

If P is a point from the circumcircle ot ABC and H is the orthocentre, prove that the Simson line from P bisects HP.

This follows from a more general configuration: Given 4 coplanar points A,B,C,D with no 3 of them collinear, then the pedal circle of each point with respect to the triangle formed by the remaining points concur at the Poncelet point of A,B,C,D. Now, if A,B,C,D are concyclic, then the pedal circle of D with respect to ABC becomes the Simson line d of D passing through the Poncelet point of the cyclic ABCD, i.e. the center of symmetry of ABCD and the quadrilateral formed by the orthocenters of BCD,CDA,DAB,ABC. As a result, d bisects the segment connecting D with the orthocenter of ABC.


LaTeX
number theory unsolved
number theory

Find all quadruple $ (m,n,p,q) \in \mathbb{Z}^4$ such that \[ p^m q^n \equal{} (p\plus{}q)^2 \plus{} 1.\]

sorry, I haven't learned to use the LATEX,so I can't post the solution.
I've just left one condition: q^n=x^2+1 have no interger solutions.

The equation is equivalent to $p^mq^n=2pq+p^2+q^2+1$. This follows $pq|p^2+q^2+1$. Let $p^2+q^2+1=kpq$. Thus, $p^2-p \cdot kq+q^2-1=0 \qquad (1)$.
Denote $(p,q)$ be the positive integer solution of this equation $(1)$ such that $p+q$ is minimum. WLOG $p \ge q$. Let the $p_0$ be the left solution. Thus, by Viete theorem we have $kq=p+p_0$ and $q^2+1=pp_0$. Since $q^2+1>0,p>0$ then $p_0>0$. This follows $(p_0,q)$ is also the positive integer solution of equation $(1)$. Hence we have $p+q \le p_0+q$ or $p \le p_0$ or $p^2 \le p_0p$ or $p^2 \le q^2+1$.
On the other hand, we also have $p \ge q$ and we can see that $p \ne q$ so $p \ge q+1$ or $p^2 \ge q^2+1+2q$, a contradiction.
Thus, there is no solution for the equation. $\blacksquare$

If $m,n > 0$ ,since $p^mq^n = (p^2+2pq+q^2)+1$ we have
\[pq | p^2+q^2+1      (I)\]
but this is a well known result that if this occurs for integer $p,q$ then $\frac {p^2+q^2+1}{pq}=3$ so $p^mq^n=5pq$ which results in $p^{m-1}q^{n-1}=5$
So either $|p|=5$ and $m=2$ or $|q|=5$ and $n=2$ , WLOG $|p|=5$ and $m=2$. If $p=-5$ then obviously $q$ has to be $-1$ so $pq=5$ and $p^2+q^2+1=27$ so according to $(I)$ $5 | 27$ which is a contradiction, so $p=5$. According to $(I)$ , $q|26+q^2$ which yields in $q|26$. So $|q| \in {1,2,13,26}$ but if $|q|=1$ or $|q|=26$ then $q^2+1 =2 mod5$ which is a contradiction. So $|q| \in {2,13}$ but in both conditions $p^{m-1}q^{n-1}=5$ yields in $n=1$ and according to the first equation $p^mq^n=(p+q)^2+1>0$ so $25q>0$ so $q>0$. so $q \in {2,13}$ which both are correct.

The solutions are $(p,q,m,n)=(5,2,2,1) , (5,13,2,1) , (13,5,1,2) , (2,5,1,2)$

[quote="Rosette.H"][quote="shinichiman"]Thus, by Viete theorem we have $kq=p+p_0$ and $q^2+1=pp_0$. [/quote]
Do you want to say $q^2-1=pp_0$ ?[/quote]
Sorry, my solution was wrong.  :(

my solution:
Lemma: $x,y$ are two integers such that $xy\mid x^2+y^2+1$ then $\frac{x^2+y^2+1}{xy}=3$.
It's well known for solution you can see [url=http://artofproblemsolving.com/community/c146h150373p849194]this[/url].
Back to the main problem:
Note that $p^2+q^2+1=(p^{m-1}q^{n-1}-2)pq$ by lemma $p^{m-1}q^{n-1}-2=3\longrightarrow p^{m-1}q^{n-1}=5$ so the solutions are:
$(m,n,p,q)=(1,2,13,5),(1,2,2,5),(2,1,5,13),(2,1,5,2)$
DONE


geometry
calculus
rectangle
MATHCOUNTS
trapezoid
analytic geometry
integration

[quote="Treething"]The lines y = 2x, x = 92, x = 99, and y = 0 are graphed. What is the area of the quadrilateral contained between these lines.

You can solve it using calculus [i]but you don't have to[/i].[/quote]

[hide]The top triangle has legs 7 and 14, area 49. The rectangle has sides 184 and 7, area $\boxed{1337}$. nice...[/hide]

Well, as seeing this is in the Mathcounts section, it shouldn't need calculus to solve it  :D

[hide]Draw the lines to see that the quadrilateral is made up of a rectangle and a triangle.  The triangle is a right triangle and has legs of 7 and 14, making the area $\frac{14\cdot7}{2}=49$.  The rectangle is 92x14, so its area is $92\times14=1288$.  Add them together to get $49+1288=\boxed{1337}$.[/hide]

Nice answer  ;)

[quote="Treething"]The lines y = 2x, x = 92, x = 99, and y = 0 are graphed. What is the area of the quadrilateral contained between these lines.

You can solve it using calculus [i]but you don't have to[/i].[/quote]

[hide]$(99+92)(99-92)=\boxed{1337}$[/hide]

[hide]I found out it was a trapezoid and it's bases and heights.  I got an answer of 1337.

$1337$[/hide]

[hide]It's a trapezoid with height 7, and bases 184 and 198. Using the formula for area of a trapezoid, you get 1337.[/hide]

Calculus just makes it ten times harder, it's not a curve or anything...

[quote="Treething"]The lines y = 2x, x = 92, x = 99, and y = 0 are graphed. What is the area of the quadrilateral contained between these lines.

You can solve it using calculus [i]but you don't have to[/i].[/quote]
[hide]This may seem stupid but you can use the shoestring. one point is (92,184) another is (99,198) another is (92,0) and the last is (99,0).[/hide]

Or we don't have to.

But is shoestring that important especially since you have the knowledge of finding the area of a figure in a coordinate plane without any complicated formulas.

At first I solved with shoestring, but then I looked at the solutiuons of others and finding the area of a trapezoid is MUCH easier.

I am not really good at calculus and I only know the basics of it.  That is why I want to only use what I used in Mathcounts to help me with what I need to know for these questions.

[hide]let's see...
calculus... you asked for it  :P
 we need to take the integral of $f(x)=2x$ from $x=92$ to $x=99$. this is written as
$\displaystyle\int^{99}_{92} 2x\,dx = F(99)-F(92)$
$F(x)=x^2$ is the antiderivative of $f(x)=2x$
therefore, $F(99)=99^2$ and $F(92)=92^2$. solving gives 1337.[/hide]

Actually we never asked for a solution with calculus, right?

:blush: yea, but you never said anything that disallowed(?) me from giving such a solution. :P

Because that is what a Mathematician is?  He/She needs to consider all cases in a problem in order to find a solution.  That is why they are literal.

[hide]Trapezoid with height 7 and bases 198 and 184. Area=1337.[/hide]


Five balls are numbered 1 through 5 and placed in a bowl.  Josh
will randomly choose a ball from the bowl, look at its number and
then put it back into the bowl.  Then Josh will again randomly
choose a ball from the bowl and look at its number.  What is the
probability that the product of the two numbers will be even and
greater than 10?  Express your answer as a common fraction.

You're supposed to input an entire solution, not a random number that could be correct (in this case it isn't) decorated with a possibly insulting remark.

Here we incorporate complementary probability, so we find a way for the product to be odd, then subtract it out because if the product isn't odd then it is even.

$ 1 \minus{} \left(\frac {3}{5}\right)^2 \equal{}\frac {16}{25}$

We still have to take into account, however, that it has to be greater than 10, so we remove the $ 9$ such possibilities, leaving $ \boxed{\frac{7}{25}}$.

The answer on FTW is 1/5.
The two numbers chosen have to be either Even/Even or Even/Odd in order to get an even product.
Therefore, each pair must have either a 2 or a 4.

The highest possible product with a two is 2*5. 10 is not greater than ten, so you must use a 4.

I guess you could easily brute force this if you know your basic times tables :P
4*1=4
4*2=8

4*3=12
4*4=16
4*5=20
Then there are the ways to mix up 4/3 and 4/5 (3/4 and 5/4). 4/4 switched would just be the same. Therefore the total number of valid ways is: 5. The total number of ways to pick two balls with replacement is 5*5, or 25.

5/25=1/5.


abstract algebra
superior algebra
superior algebra unsolved

Let $ p$ an odd prime and $ \mathbb{F}_{p^n}$ the finite field with $ p^n$ elements.

$ i)$ Prove that the set $ \mathbb{F}^2 = \{a^2, \ a\in\mathbb{F} \}$ has $ \frac {p^n + 1}{2}$ elements. Then, if $ t\in\mathbb{F}$ then the set $ t - \mathbb{F}^2 = \{ t - a^2, a\in\mathbb{F} \}$ has $ \frac {p^n + 1}{2}$ elements.

$ ii)$ For $ t\in\mathbb{F}$ prove that the set $ F^2\cap(t - \mathbb{F}^2)$ is not the empty set and then prove that every element $ c\in\mathbb{F}$ can be written in the form $ c = a^2 + b^2, \ \ a,b\in \mathbb{F}$.

$ iii)$ Prove that the equation $ x^2 + y^2 + z^2 = 0$ has nontrivial solution in $ \mathbb{F}$ (A trivial solution is the $ x = y = z = 0$).

Alexandros

a. the restriction of the map to nonzero elements is a homomorphism with respect to multiplication and the kernel is 1 and -1, thus the image has $ \frac{p^n\minus{}1}{2}$ elements. Together with 0, we have the desired result.

b. (I suppose you mean the intersection of $ F^2$ and $ t\minus{}F^2$)If the intersection is empty there are at least $ p^n\plus{}1$ elements in the field, which is absurd. Since $ c \minus{} F^2$ intersects $ F^2$, $ c \minus{} a^2 \equal{} b^2$ for some $ a,b$, thus $ c \equal{} a^2\plus{}b^2$ for some $ a,b$.

c. Pick nonzero $ x$, set $ c \equal{} \minus{}x^2$, and apply b.

You definitely need to spend more time proofreading- all three of your questions have errors.
#1: That set is not a field.
#2: Already pointed out.
#3: "no trivial solution" should be "a nontrivial solution"

Thanks a lot both of you! I have already edited my post! I am sorry for those misprints.

Alexandros


search
number theory proposed
number theory

[color=blue][u][b]Bertrand 's postulate[/b][/u][/color]
  Prove that: For every  $ n \ge 1$ , there is some prime number $ p$ with $ n < p \le 2n$ .
-------
  Is there always a prime between $ n^2$ and $ (n \plus{} 1)^2$ ?
 :D

Or even just [url=http://en.wikipedia.org/wiki/Proof_of_Bertrand%27s_postulate]regular search[/url].  The second question is open.