Find all real roots of the equation:
\[ \sqrt{x^{2}-p}+2\sqrt{x^{2}-1}=x\]
where $ p$ is a real parameter.

I got $ x=\frac{(p-4)(\pm \sqrt{4-2p})}{8-4p}$, with $ p < 2$.  If $ p\ge2$ there's no solution.

[quote="santosguzella"]Find all real roots of the equation:
\[ \sqrt{x^{2}-p}+2\sqrt{x^{2}-1}=x \]
where $ p$ is a real parameter.[/quote]

Notice we need $ x\geq 0$, $ x^{2}\geq p$ and  $ x^{2}\geq 1$
Then
$ \sqrt{x^{2}-p}+2\sqrt{x^{2}-1}=x$ $ \implies$ $ 2\sqrt{x^{2}-1}=x-\sqrt{x^{2}-p}$ $ \implies$ $ 4x^{2}-4=2x^{2}-p-2x \sqrt{x^{2}-p}$
$ \implies$ $ 2x^{2}+p-4=-2x \sqrt{x^{2}-p}$ $ \implies$ $ 4x^{4}+(p-4)^{2}+4x^{2}(p-4)=4x^{2}(x^{2}-p)$
$ \implies$ $ (p-4)^{2}+4x^{2}(2p-4)=0$ $ \implies$ $ p< 2$ and $ x^{2}=\frac{(p-4)^{2}}{8(2-p)}$

Now, if we put back this value in the initial equation (since we used $ \implies$ and not $ \Leftrightarrow$), we get :
$ x^{2}-1=\frac{p^{2}}{8(2-p)}$, $ x^{2}-p=\frac{(3p-4)^{2}}{8(2-p)}$, and then :
$ |3p-4|+2|p|=|p-4|$ and $ p< 2$. This equation implies $ 0\leq p\leq\frac{4}{3}$

So the solution is $ x=\frac{4-p}{\sqrt{8(2-p)}}$ with any $ 0\leq p\leq\frac{4}{3}$
which is Diogene's solution.


Find the number of cubic centimeters in the volume of the cylinder formed by rotating a square with side length 14 centimeters about its vertical, line of symmetry. Express your answer in terms of $ \pi$.

Notice that the radius is $ \frac {14}{2} \equal{} 7$ and the height is just $ 14$. Therefore, the volume is $ 7^2 \pi \cdot 14 \equal{} \boxed{686 \pi}$


quadratics
number theory
prime numbers
number theory unsolved

$ p.q$ are  prime numbers and $ p\equal{}2q\plus{}1$ and $ a\not\equiv\\minus{}1,0,1$ Show that $ \minus{}a^2$ is primitive root (mod p)

[quote="wiosna"]$ p$, $ q$ are prime numbers such that $ p \equal{} 2q \plus{} 1$ and $ a\not\equiv\ \minus{} 1,0,1$ Show that $ \minus{} a^2$ is a primitive root ($ \mod p$)[/quote]
If $ q\equal{}2$ there is nothing to show. In the sequel, let $ q \ge 3$. Hence $ p \equiv 3(\mod p)$. Therefore, $ \minus{}a^2$ isn't a quadratic residue modulo $ p$. Suppose that $ \minus{}a^2$ is no primitve root. Let $ g$ be a primitiv root modulo $ p$ (as $ p$ is a prime, $ g$ exists). Hence there exists an $ l\ge 1$ such that $ g^{2l\plus{}1}\equal{}\minus{}a^2$ (this is because $ \minus{}a^2$ is no quadratic residue). Since $ \minus{}a^2$ is no primitive root, there exists a $ k<p\minus{}1$ such that $ g^{(2l\plus{}1)k}\equal{}(\minus{}a^2)^k \equiv 1$ $ (\mod p)$. This, together with Fermat's little theroem, implies $ (2l\plus{}1)k | p\minus{}1\equal{}2q$ and hence $ k\equal{}2$. Therefore, $ a^4 \equiv 1 (\mod p)$ and by Fermat's little theorem $ 4|2q$, contradiction. Hence $ \minus{}a^2$ is a primitive root.


$20$ students participated at a table tennis contest that lasted for $3$ days. Every day, every player played exactly one match (it is possible that the same pair plays more than one match). When the contest was over, it was noticed that there is only one winner and that none of the players lost all $3$ matches.

How many students won exactly one match? How many two matches?

[i]G. Rene[/i]

[hide]Table tennis matches can't tie.

There are $\frac{20}{2}\cdot 3=30$ total matches, and thus $30$ total wins, played during the tournament. There is only one winner, so exactly one person got three wins. (The winner couldn't have won two matches, because otherwise everyone else would have won one match and the total number of matches would be $21$, which is not $30$.) This leaves $27$ match wins and $19$ people, some of which won $1$ match and the rest of which won $2$ matches. Solving this simple system of equations gives that

$8$ people won two matches, and
$11$ people won one match.

(It's probably also easy to show that this result it attainable)[/hide]

[hide="solution"]
11 people won 1 game and 8 people won 2.  For every person that wins 2 games there is someone that loses 2, or wins 1 in other terms.  For the person who wins 3 games there are two who win only 1 game.

[/hide]


a, b, c, d, and e are consecutive positive integers.  The sum of b, c, and d is a perfect square, and the sum of a, b, c, d, and e is a perfect cube.  What is the lowest possible value of c.

So can you solve this using a matrix?  If not, then how would you go about solving it?  Sorry if this is an easy problem, I'm new to AIME.

What do you mean by "using a matrix"?  It's a number theory question and has a number theory answer.

[hide="solution without matrix"]
$ (a,b,c,d,e)\equal{}(c\minus{}2,c\minus{}1,c,c\plus{}1,c\plus{}2)$,then with positive integers $ m,n$ we have

$ (c\minus{}1)\plus{}c\plus{}(c\plus{}1)\equal{}3c\equal{}m^2$ and $ (c\minus{}2)\plus{}(c\minus{}1)\plus{}c\plus{}(c\plus{}1)\plus{}(c\plus{}2)\equal{}5c\equal{}n^3$ hence

$ 5m^2\equal{}3n^3$,since $ GCD(5,3)\equal{}1$,we have $ 3|m^2$ and $ 5|n^3$ so $ 3|m$ and $ 5|n$,

therefore let $ m\equal{}3M,n\equal{}5N$ we have $ 3M^2\equal{}25N^2$ so $ 3|N^2\Longrightarrow 3|N\Longrightarrow N\ge 3\Longrightarrow c\equal{}25N^3\ge 25\times 3^3\equal{}675$.

equality holds if $ (m,n)\equal{}(45,15)$,hence the least $ c$ is $ 675$.[/hide]


Let $ a_1,a_2,...,a_n$ be positive real numbers,and $ n\in N, n\geq 2$, such that $ a_1\plus{}a_2\plus{}\cdots\plus{}a_n\equal{}n$.
Prove that the following inequality holds

$ \frac{1\minus{}a^2_1}{1\plus{}a^2_1}\plus{}\frac{1\minus{}a^2_2}{1\plus{}a^2_2}\plus{}\cdots\plus{}\frac{1\minus{}a^2_n}{1\plus{}a^2_n}\geq 0$

It is a bit easy . Rewrite it as:

\[ \frac{n}{2}\ge \sum_{i\equal{}1}^n\frac{a_i^2}{1\plus{}a_i^2}\]

which is obvious by AM-GM.

[quote="marin.bancos"]Let $ a_1,a_2,...,a_n$ be positive real numbers,and $ n\in N, n\geq 2$, such that $ a_1 \plus{} a_2 \plus{} \cdots \plus{} a_n \equal{} n$.
Prove that the following inequality holds

$ \frac {1 \minus{} a^2_1}{1 \plus{} a^2_1} \plus{} \frac {1 \minus{} a^2_2}{1 \plus{} a^2_2} \plus{} \cdots \plus{} \frac {1 \minus{} a^2_n}{1 \plus{} a^2_n}\geq 0$[/quote]

Thank you marin bancos for your inequality  :lol:

From Am-Gm inequality

$ \frac {1 \minus{} a^2_1}{1 \plus{} a^2_1} \equal{} 1 \minus{} \frac {2a_1^2}{a_1^2 \plus{} 1} \ge 1 \minus{} a_1$


Now something is completely wrong with my TI-89 titanium.  I downloaded Pokemon( :D). I ran the program and in capital letters, at the top it said "ADRESS ERROR".  Now i cant turn it off, i can't press any buttons, i cant do anything!!!!!!!  And i just got this calculator 2 days ago and its all messed up.  Can someone tell me what to do!?!?!?!??!?!?!??!?!?!?!?!??!!!!!!!!!!!!!?????????????????!!!!!!!!!!!!!!
!!!!!??????????????????!!!!!!!!!!!!!!!!!!????????????!!!!!!!!!!!!!!!??????
?????

Perhaps you could try pressing the "ON" button, since this should stop a program's execution.  If that doesn't work, you could also try taking the batteries out and putting them back in again as a last resort, but do this at your own risk since you could lose any stored information on your calculator...

ok i fixed it...phew...pressing on didn't do anything, and i had to take out the back up battery with a screwdriver!

2nd + Left + Right + On = hard reset.  Wipes out RAM, but at least your calculator works again.


trigonometry
logarithms
real analysis
real analysis solved

Prove that the sequence $\displaystyle (a_{n})_{n \geq 1}$ with $a_{n}=\cos (\ln(n))$ does not converge.

Since $\ln {n+1}-\ln n\to 0$ as $n\to \infty$, it means that when we place the points $A_n$ representing $\ln n$ on the real line we will fall in an interval of the form $(2k\pi+a,2k\pi+b),\ b<2\pi$ infinitely many times, so the sequence $\{\ln n\}$ is dense in $(0,2\pi)$, where the fractional part is taken $\mod 2\pi$ instead of $\mod 1$ as usual. This means that for any numbers $a,b\in (0,\pi)$ there are infinitely many $n$ for which $\cos{\ln n}\in (\cos b,\cos a)$ and we're done.


geometry
circumcircle
cyclic quadrilateral
geometry proposed

Let $ABCD$ be a convex quadrilateral, and call $R$ the intersection of its diagonals. Now, call $\omega$ the circumcircle of $\Delta ABR$, and draw lines $l_{1}$ and $l_{2}$, tangent to $\omega$ at $A$ and $B$ respectively. We call $l_{1}\cup l_{2}= \left\{S\right\}$

If we call $M$ the midpoint of $\left[CD\right]$, then prove:
$M,R,S$ are collinear $\iff$ $ABCD$ is a cyclic quadrilateral.

Please tell me whatever you think about this problem.

$($ In my drawing $AB = 6.2,$ $BC = 7.0,$ $CD = 8.0,$ $R = 6.0,$ where $R,$ is the circumradius of the cyclic quadrilateral $ABCD$ $).$

[b][size=117]1.[/size][/b] – We suppose that the quadrilater $ABCD,$ is cyclic. So, if we consider the circumcircle $(O),$ of $ABCD,$ as an arbitrary circle taken as cord, the sideline $AB,$ of the triangle $\bigtriangleup ABR,$ then the segment line $CD$ $($ where $C,$ $D,$ are considered as the intersection points of the sidelines $AR,$ $BR,$ of $\bigtriangleup ABR$ from the circle $(O)$ $),$ is antiparallel to $AB,$ with respect to the sidelines $AR,$ $BR,$ of the angle $\angle ARB.$

So, the segment line $MR,$ where $M,$ is the midpoint of the segment $CD,$ is the symmedian of $\bigtriangleup ABR,$ through it’s vertex $R$ and as well known, passes through the point $S,$ as the intersection point of the tangent lines of $\bigtriangleup ABR,$ at it’s vertices $A,$ $B.$

Hence, we conclude that the points $M,$ $R,$ $S,$ are collinear.

[b][size=117]2.[/size][/b] – We suppose now, that the points $M,$ $R,$ $S,$ are collinear and the quadrilateral $ABCD,$ is not cyclic. So, we draw the circumcircle of the triangle $\bigtriangleup ABC,$ which intersects the segment line $BR,$ at a point so be it, $D'.$

We denote as $M',$ the midpoint of the segment $CD'$ and based on all the above, we conclude that the points $M',$ $R,$ $S,$ are collinear and hence we have that $MM'\parallel DD'$ $($ because $M,$ is also the midpoint of the segment $CD$ $).$

But this result is [b][size=109]''atopo''[/size][/b] $($ = improper $),$ because $R\equiv MM'\cap DD'.$ So, we conclude that $D'\equiv D.$ Hence, the quadrilateral $ABCD,$ is cyclic and the proof is completed.

Kostas Vittas,

Thank you for your proof Mr. Kostas Vittas.
I hope you enjoyed the problem  :)


[quote="guru_bhai"]1) Let $ p(x) \equal{} x^{10} \plus{} a_1x^9 \plus{} ....... \plus{} a_{10}$ be a polynomial with real coefficients. Suppose $ p(0) \equal{} \minus{} 1, p(1) \equal{} 1, p(2) \equal{} \minus{} 1$. Let $ R$ be the number of real zeros of $ p(x)$.Which of the following must be true ?
$ (A) R \ge 4$
$ (B) R \equal{} 3$
$ (C) R \equal{} 2$
$ (D) R \equal{} 1$

2) Let $ p(x)$ and $ q(x)$ be non-zero polynomials with real coefficients such that degree$ (p(x)) >$degree$ (q(x))$. If the graphs of $ y \equal{} p(x)$ and $ y \equal{} q(x)$ intersect in 3 points, which of the following must be true ?
$ (A)$ degree$ (p(x)) \ge 2$
$ (B)$ degree$ (p(x)) \ge 3$
$ (C)$ degree$ (p(x)) \equal{} 2$
$ (D)$ degree$ (p(x)) \equal{} 6$[/quote]
First of all this problem should not be in olympiad section.

for 1.
it is clear that $ p$ has 2 real zeroes.Also as $ x\to \infty,p\to \infty$ so A is correct.
2.B
because $ p\minus{}q$ has at least three real roots.

How did you deduce that $ p$ has 2 real zeros? I could only see $ p \plus{} 1$ has at least two real zeros...

As $ p$ ranges from $ 0$ to $ 2$, it starts under the $ x$-axis, goes above it, then goes back under. What ith_power is saying is that as $ x$ gets smaller and smaller or larger and larger, it will go back over the $ x$-axis again. Thus it crosses the $ x$-axis at least $ 4$ times.


How do I get the outside parenthesis to be larger than the inside ones? I tried using [code] \displaystyle \left((a+b)(a-b) \right) [/code], but it didn't work :blush:

\[ \left((a+b)(a-b) \right) \]

[code]$\left(a^2(a+b)+b^2\right)$[/code]

$\left(a^{2}(a+b)+b^{2}\right)$

It also works for floor, ceiling, absolute value, etc.

What if it has to be within [code] \[  \] [/code], since I'm trying to center it?

\[\left(a^{2}(a+b)+b^{2}\right) \]

It works like that, too.

[quote="Fraxia"]How do I get the outside parenthesis to be larger than the inside ones? I tried using [code] \displaystyle \left((a+b)(a-b) \right) [/code], but it didn't work :blush:
\[\left((a+b)(a-b) \right) \]
[/quote]As mysmartmouth says \left( and \right( work well when the inside is larger than the normal () (if you have a square or a fraction for example) but if you have $((a+b)(a-b))$ the outside () don't need to be larger to cope with what's inside. To get over this you can insist on the outside () being larger by using things like \biggl(. There are 4 sizes \big, \Big, bigg and \Bigg so you can do mad things like
\Biggl(\biggl(\Bigl(\bigl((a+b)(a-b)\bigr)\Bigr)\biggr)\Biggr)
which gives
\[\Biggl(\biggl(\Bigl(\bigl((a+b)(a-b)\bigr)\Bigr)\biggr)\Biggr) \]
Again you can do the same with [ | { etc

That's interesting.  I've never seen it before.  Is that an AoPS limitation to keep server load down, or is it an actual LaTeX thing?


vector
analytic geometry
function
linear algebra
matrix
category theory
articles

What are tensors, and how are they used? Thanks.

quanitities like speed in which scalar and vectors are thoroughly atrocited :P  :)

Thanks for that thurough answer rt_08  :P . Anyone care to add onto what he said?

Hmm, I've been trying to understand tensors (without too much success), but here's what I've gathered.

If we have a vector, then for a given basis, there is a set of numbers that are that vector's coordinates with respect to that basis. For example, in the one-dimensional case, if we have a basis consisting of the vector $ e$, then the coordinates of the vector $ x \equal{} 5e$ consist simply of the number 5.

Now, if we know the coordinates of the vector in a particular basis (say $ \{a\}$), that should determine which vector it is (as an element of the vector space), and so we should be able to determine its coordinates in any other basis. Specifically, if we know a vector's coordinates in basis A, how do we determine its coordinates in basis B? This is the so-called "transformation of coordinates." For example, if $ b \equal{} 2a$ is another basis, then $ x \equal{} 5a \equal{} \frac {5b}{2}$, so the coordinate of $ x$ is $ \frac {5}{2}$.

Now think about linear functionals on a vector space, i.e. linear functions from a vector space to the field of scalars. If we have a linear functional f, and we have a basis, then f is determined by what it does to each element of the basis. We write the "coordinates" of this linear functional by putting in a coordinate for each basis element that equals f of that basis element. Let's consider a function f such that $ f(a) \equal{} 5$, so that its representation with respect to that basis is the coordinate 5. One way to think of f is that for any vector, it takes that vector's coordinate with respect to a, and then multiplies that number by 3. Now what happens if we try to express this function in the basis $ b$? We have $ f(b) \equal{} f(2a) \equal{} 2f(a) \equal{} (2)(5) \equal{} 10$. So, the coordinate of f in b is 10, not $ \frac {5}{2}$, and so we see a key idea, that the coordinates of these functions transform differently under change of basis than those of vectors.

Now what else has a representation with respect to a particular basis? Well, linear functions from the space to itself do (called "linear operators"). For a particular such function, say m, and a given basis, we can associate a matrix of coordinates. For a different basis, but for the SAME underlying function, we have a different set of numbers (for an n-dimensional vector space, this consists of $ n^2$ numbers). Now, how do these transform, i.e. if we're given two bases, and we know the coordinates of the function with respect to one basis, how do we find the coordinates of the function with respect to the other basis? It turns out that the transformation rule, although slightly more complicated, sort of has elements of both the "function" and the "vector" rules. In fact, in the one-dimensional case, these two cancel, and a function's representation is independent of basis.

So I've been talking about things that have an underlying "structure," such as a function or vector, and then are associated with a specific set of numbers for each basis. A way to think of these things is as equivalence classes of pairs, each consisting of a basis and a set of numbers (the coordinates). In the case of vectors, these equivalence classes of bases and sets of numbers (in the case of an $ n$-dimensional vector space, $ n$ numbers) each correspond to an underlying element of the vector space. For a function (both functionals and function to the space itself), each equivalence class represents an underlying function on the space of elements.

Now let me take the time to give an example which is less studied than vectors, linear operators, or linear functionals, and that is, bilinear functionals, that is, functions g in two arguments that take values in the scalars such that if we fix one argument, we get a linear function of the other. It turns out that we can determine the function by what it does to pairs of basis elements, and so it can in fact be determined by a set of $ n^2$ coordinates (for an $ n$-dimensional vector space). So, like linear operators, it has $ n^2$, meaning it's isomorphic as a vector space (note: we can make these functions into a vector space), so, I mean, do its coordinates change in the same way under coordinate change? The answer is no. In reality, it has a twofold "function" nature, and so it transforms even more in the "function" way. This is what makes them distinct from the linear operators, not the vector space structure, but these different sets of coordinates with respect to different bases, this additional structure to the vector space structure. Another way to look at it, it's which equivalence relation you choose that makes them fundamentally different.

(In fact, you should have already noticed that the space of functionals [the "dual space"] and the space of vectors are isomorphic as vector spaces [only in the finite dimensional case, actually - but don't worry about that fact right now])

Now, I've said a lot about this concept of different sets of coordinates with respect to different bases, and I think you've gotten the gist of it. As you can figure, everything I've told you about are all examples of "tensors." Now, for a tensor in general, we just consider higher-order sets of coordinates (say $ n^3$ or more) and define transformation rules similar to the ones mentioned. In fact, a space of tensors in general is defined by two numbers, $ (i,j)$, and it has $ n^{i \plus{} j}$ coordinates. This essentially means that it has $ i$ "functionness" and $ j$ "vectorness" in its transformation rules. The space of vectors is (0,1), the space of functionals (1,0), the space of linear operators is (1,1), and the space of bilinear functionals is (2,0). Higher order, say (3,4), or whatever, is just a generalization of this concept of a different set of coordinates with respect to each basis, and they transform (i.e. which set of coordinates in basis A corresponds to which set of coordinates in basis B) in a certain way.

One note: covariant and contravariant are the proper terms for what I called "function" and "vector" transformation.

Also, I'm sorry, but I don't know much about applications to physics, although this is very relevant when learning about tensors!

Check out Wikipedia; it has some good articles on the subject.

A tensor is an object which extends the notion of scalar, vector, and matrix. The term has slightly different meanings in mathematics and physics. In the mathematical fields of multilinear algebra and differential geometry, a tensor is a multilinear function. In physics and engineering, the same term usually means what a mathematician would call a tensor field: an association of a different (mathematical) tensor with each point of a geometric space, varying continuously with position.

Mathematical
In mathematics, a tensor is (in an informal sense) a generalized linear 'quantity' or 'geometrical entity' that can be expressed as a multi-dimensional array relative to a choice of basis of the particular  :)  :P  space on which it is defined. The intuition underlying the tensor concept is inherently geometrical: as an object in and of itself, a tensor is independent of any chosen frame of reference. However, in the modern treatment, tensor theory is best regarded as a topic in multilinear algebra. Engineering applications do not usually require the full, general theory, but theoretical physics now does.

For example, the Euclidean inner product (dot product)- a real-valued function of two vectors that is linear in each—is a mathematical tensor. Similarly, on a smooth curved surface such as a torus, the metric tensor (field) essentially defines a different inner product of tangent vectors at each point of the surface. Just as a linear transformation can be represented as a matrix of numbers with respect to given vector bases, so a tensor can be written as an organized collection of numbers. In physics, the numbers may be obtained as physical quantities that depend on a basis, and the collection is determined to be a tensor if the quantities transform appropriately under change of basis.

 Physical - tensor fields
Many mathematical structures informally called 'tensors' are actually tensor fields—a tensor valued function defined on a geometric or topological space. This use of the term is analogous to vector fields such as electromagnetic fields, but with the 'tensor' defined so that it is invariant under a change of coordinates. Differential equations posed in terms of tensor quantities are basic to modern mathematical physics, so that tensor fields are usually defined on differentiable manifolds.

 Tensor rank
In mathematics, the term rank of a tensor may mean either of two things, and it is not always clear from the context which.

In the first definition, the rank of a tensor T is the number of indices required to write down the components of T. This is the sum of the number of covariant and contravariant indices. Expressed by means of the tensor product of multilinear algebra, this is the number of factors of the tensor product needed to express T.

In the second definition, the rank of a tensor is defined in a way that extends the definition of the rank of a matrix given in linear algebra. A tensor of rank 1 (also called a simple tensor) is a tensor that can be written as a tensor product of the form

where a, b,...,d are in V or V*. In indices, a tensor of rank 1 is a tensor of the form

Every tensor can be expressed as a linear combination of rank 1 tensors. In general, the rank of T is the minimum number of rank 1 tensors with which it is possible to express T as a linear combination.

For example, a matrix is a tensor with 2 indices, and so has rank 2 in the first definition. On the other hand, the rank of the tensor in the second definition is just the rank of the matrix. This latter meaning is possibly the intended one, whenever the array of components is two-dimensional.

To avoid this ambiguity, it is now preferred to use the terminology of tensor order to denote the number of indices, and tensor rank to designate the number of simple tensors necessary to decompose a tensor. Hence the definition of rank is now used in a way that is consistent with Linear Algebra.

Tensor valence
In physical applications, array indices are distinguished by being contravariant (superscripts) or covariant (subscripts), depending upon the type of transformation properties. The valence of a particular tensor is the number and type of array indices; tensors with the same tensor order but different valence are not, in general, identical. However, any given covariant index can be transformed into a contravariant one, and vice versa, by applying the metric tensor. This operation is generally known as raising or lowering indices.

Importance and applications
Tensors are important in physics and engineering. In the field of diffusion tensor imaging, for instance, a tensor quantity that expresses the differential permeability of organs to water in varying directions is used to produce scans of the brain; in this technique tensors are in effect made visible. Perhaps the most important engineering examples are the stress tensor and strain tensor, which are both 2nd rank tensors, and are related in a general linear elastic material by a fourth-rank elasticity tensor.

Specifically, a 2nd rank tensor quantifying stress in a 3-dimensional/solid object has components which can be conveniently represented as a 3x3 array. The three Cartesian faces of a cube-shaped infinitesimal volume segment of the solid are each subject to some given force. The force's vector components are also three in number (being in three-space). Thus, 3x3, or 9 components are required to describe the stress at this cube-shaped infinitesimal segment (which may now be treated as a point). Within the bounds of this solid is a whole mass of varying stress quantities, each requiring 9 quantities to describe. Thus, the need for a 2nd order tensor is produced.

While tensors can be represented by multi-dimensional arrays of components, the point of having a tensor theory is to explain further implications of saying that a quantity is a tensor, beyond specifying that it requires a number of indexed components. In particular, tensors behave in specific ways under coordinate transformations. The abstract theory of tensors is a branch of linear algebra, now called multilinear algebra.

The choice of approach
There are two ways of approaching the definition of tensors:

The usual physics way of defining tensors, in terms of objects whose components transform according to certain rules, introducing the ideas of covariant or contravariant transformations.
The usual mathematics way, which involves defining certain vector spaces and not fixing any coordinate systems until bases are introduced when needed. Contravariant vectors, for instance, can also be described as one-forms, or as the elements of the dual space to the covariant vectors.
Physicists and engineers are among the first to recognise that vectors and tensors have a physical significance as entities, which goes beyond the (often arbitrary) coordinate system in which their components are enumerated. Similarly, mathematicians find there are some tensor relations which are more conveniently derived in a coordinate notation.

 Physical examples
As a simple example, consider a ship in the water. We want to describe its response to an applied force. Force is a vector, and the ship will respond with an acceleration, which is also a vector. The relationship between force and acceleration is linear in classical mechanics. Such a relationship is described by a rank two tensor of type (1,1) (that is to say, here it transforms a plane vector into another such vector). The tensor can be represented as a matrix which when multiplied by a vector results in another vector. Just as the numbers which represent a vector will change if one changes the coordinate system, the numbers in the matrix that represents the tensor will also change when the coordinate system is changed.

In engineering, the stresses inside a solid body or fluid are also described by a tensor; the word "tensor" is Latin for something that stretches, i.e., causes tension. If a particular surface element inside the material is singled out, the material on one side of the surface will apply a force on the other side. In general, this force will not be orthogonal to the surface, but it will depend on the orientation of the surface in a linear manner. This is described by a tensor of type (2,0), in linear elasticity, or more precisely by a tensor field of type (2,0) since the stresses may change from point to point.

Mathematical examples
Some well-known examples of tensors in differential geometry are quadratic forms, such as metric tensors, and the curvature tensor.

Formally speaking, a tensor has a particular type according to the construction with tensor products that give rise to it. For computational purposes, it may be expressed as the sequence of values represented by a function with a tuple-valued domain and a scalar valued range. Domain values are tuples of counting numbers, and these numbers are called indices. For example, a rank 3 tensor might have dimensions 2, 5, and 7. Here, the indices range from «1, 1, 1» through «2, 5, 7»; thus the tensor would have one value at «1, 1, 1», another at «1, 1, 2», and so on for a total of 70 values. As a special case, (finite-dimensional) vectors may be expressed as a sequence of values represented by a function with a scalar valued domain and a scalar valued range; the number of distinct indices is the dimension of the vector. Using this approach, the rank 3 tensor of dimension (2,5,7) can be represented as a 3-dimensional array of size 2 × 5 × 7. In this usage, the number of "dimensions" comprising the array is equivalent to the "rank" of the tensor, and the dimensions of the tensor are equivalent to the "size" of each array dimension.

A tensor field associates a tensor value with every point on a manifold. Thus, instead of simply having 70 values as indicated in the above example, for a rank 3 tensor field with dimensions «2, 5, 7»; every point in the space would have 70 values associated with it. In other words, a tensor field means there's some tensor-valued function which has, for example, Euclidean space as its domain.

 Approaches, in detail
There are equivalent approaches to visualizing and working with tensors; that the content is actually the same may only become apparent with some familiarity with the material.

The classical approach
The classical approach defines a tensor to a collection of multidimensional arrays, such that one array is associated to each possible coordinate system of any fixed vector space. This notion generalizes scalars, vectors, matrices, linear functionals, bilinear forms, etc. To represent a vector x as a tensor one can simply let the array associated to any basis B be the vector of coordinates of x with respect to B.
However, to count as a tensor, the arrays need to obey a relation that precisely corresponds to how vectors, matrices, linear functionals, etc transform when one passes from one coordinate system to another.
The modern approach
The modern (component-free) approach views tensors initially as abstract objects, expressing some definite type of multi-linear concept. Their well-known properties can be derived from their definitions, as linear maps or more generally; and the rules for manipulations of tensors arise as an extension of linear algebra to multilinear algebra. This treatment has attempted to replace the component-based treatment for advanced study, in the way that the more modern component-free treatment of vectors replaces the traditional component-based treatment after the component-based treatment has been used to provide an elementary motivation for the concept of a vector. One could say that the slogan is 'tensors are elements of some tensor space'. Nevertheless, a component-free approach has not become fully popular, owing to the difficulties involved with giving a geometrical interpretation to higher-rank tensors.
The intermediate treatment of tensors attempts to bridge the two extremes, and to show their relationships.
In the end the same computational content is expressed. See glossary of tensor theory for a listing of technical terms.

 Tensor densities
Main article: Tensor density
It is also possible for a tensor field to have a "density". A tensor with density r transforms as an ordinary tensor under coordinate transformations, except that it is also multiplied by the determinant of the Jacobian to the rth power. Invariantly, in the language of multilinear algebra, one can think of tensor densities as multilinear maps taking their values in the (1-dimensional) space of n-forms (where n is the dimension of the space), as opposed to taking their values in just R. Higher "weights" then just correspond to taking additional tensor products with this space in the range. In the language of vector bundles, the determinant bundle of the tangent bundle is a line bundle that can be used to 'twist' other bundles r times.

i think that would do!
 :P

So for the moment of inertia tensor, which is a (1 1) tensor, why can't we just view it as a linear transformation? I mean, isn't that basically what a (1 1) tensor is?

Also, when tensors are used in general relativity, I heard it's just a (0 2) (I hope I have the order correct) tensor, i.e. a bilinear form that gives the metric. Is there really any need for referring to tensors? I mean, I already understand the concept of inner product / metric, so does general relativity really need to refer to the concept of "tensors"?

hey look whenever we consider general realitivity all the objects are viewed in a different perspective coz all the fundamenatal rules dont apply there!


Let a,b,c are positive real numbers.Prove that or disprove that
$ \sum\sqrt{\frac{a^{2}+ab+b^{2}}{c^{2}+ab}}\geq\frac{3\sqrt{6}}{2}$

I'm not sure how that sum would work but the problem seems easy to disprove. Let $ a=1$, $ b=2$ and $ c=9$.

$ \sqrt{\frac{1+2+4}{81+2}}$ clearly isn't greater than $ \frac{3\sqrt{6}}{2}$. If this is wrong, it probably has something to do with the sum, and if so please clarify the problem a little more. Usually, a sum is written in the following form:

$ \sum_{a=1}^{b}a^{2}+1$ where we evaluate $ a^{2}+1$ from values $ 1$ to $ b$ and add the resultant values together.

Prime factorization, PHONGVAN has meant the cyclic sum. :wink:


wat is the specality of the number 1729?????????

someone should know this.! :wink:

And it's the number of a cab that GH Hardy took to visit Ramanujan.

hello, see here
http://mathworld.wolfram.com/1729.html
Sonnhard.

That's like the least number that can be written as the sum of two squares right?

Two cubes in 2 different ways, I think.

-rts2007

Yes, as it mentions in the Wolfram link ($ 12^3\plus{}1^3, 9^3\plus{}10^3$).

Now go find the smallest number that can be written as the sum of two squares in two different ways.

I never said the squares have to be postive  :wink: . $ 25 \equal{} 5^2 \plus{} 0^2 \equal{} 3^2 \plus{} 4^2$.

However, if they must be positive, 50 is indeed the smallest.

EDIT: Well I guess I didn't say anything about the squares at all, so technically they could be squares of irrational numbers  :P .

[quote="FantasyLover"][quote="xpmath"]However, if they must be positive, 50 is indeed the smallest.
[/quote]
I believe you mean 1729.[/quote]

[quote="xpmath"]Now go find the smallest number that can be written as the sum of two squares in two different ways.[/quote]

nice fanfan

@nick42: I think you're doing squares. We're talking about cubes. (I think...)  :)

Well, 1729 is definitely not the smallest number that can be written as the sum of two squares. The smallest one is 50 and plus, $ 6^2\plus{}8^2\equal{}100$

to clear up confusion:

the original thread dealt with cubes. however, xpmath raised the question "what is the smallest number that can be expressed as a sum of two squares in moer than one way?" therefore, we are talking about both: 1729 is the smallest number that can be expressed as a sum of two CUBES in more than one way, and 50 is the smallest number that can be expressed as a sum of two SQUARES in more than one way.

hope that clears some stuff up  :)


$ x$ , $ y$ , $ z$ $ \in$ $ ] 0 , \plus{}\infty [$ such that $ x\plus{}y\plus{}z\equal{}1$ ; find minimum of :

$ \frac{x^2}{x\plus{}yz}\plus{}\frac{y^2}{y\plus{}zx}\plus{}\frac{z^2}{z\plus{}xy}$

Aply CBS :$ \sum {\frac{x^2}{x\plus{}yz}}\geq\frac{(\sum x)^2}{\sum x\plus{}\sum {yz}}\equal{}\frac{1}{1\plus{}\sum {yz}}\geq \frac{1}{1\plus{}\frac{(x\plus{}y\plus{}x)^2}{3}}\equal{}\frac{3}{4}$


[quote="mr.danh"]Let a non-equilateral triangle $ ABC$ and $ AD,BE,CF$ are its altitudes. On the rays $ AD,BE,CF,$ respectively, let $ A',B',C'$ such that $ \frac {AA'}{AD} \equal{} \frac {BB'}{BE} \equal{} \frac {CC'}{CF} \equal{} k$. Find all values of $ k$ such that $ \triangle A'B'C'\sim\triangle ABC$ for any non-triangle $ ABC.$[/quote]

Non-triangle? :P

Anyway it's problem 2 of the 2nd MathLinks Contest, 3rd Round. As I cannot find the proposed solutions on MathLinks now, I'm enclosing them here - with many thanks to Valentin and the contest participants for writing them up.

  darij


geometry
geometric transformation
rotation
vector
trigonometry

This is a dumb question.......

If we release a ballon(Those big ones with people inside),afetr a while would it land at the same place we released it from?Do not consider the air to blow it around.

I don't see a reason why should it land at any other place but perhaps I'm forgetting something... Yes, I'm forgetting the Coriolis force - it would move the balloon a little bit. So I'd say it lands at a different spot.

What's Cariolis force?

And i hope this has nothing to do witrh the concept of why a satellite tries to fall on the same point?

How about this dumb thought...
[hide]The earth rotates...so... :blush: [/hide]

That's exactly what Coriolis force is all about - the rotation of Earth. Hey, you came up with the right answer yourself! :cool:

To find out more about it, try Google. Basically, the problem with Earth's rotation is that things on Earth's surface are in fact moving in circles but these circles don't have their centers in Earth's center of gravity so the resulting centrifugal force is not parallel to the gravitational force.

Basically, if you were spinning around in a merry-go-round, and you threw a ball radially outwards, it would no longer be pulled by the spinning-ness of the merry-go-round and whatever forces it exerts on the ball.  So although from an outsider's point of view it goes straight, to the people riding on the horsies the ball curves

Hmm ... I'm not wxactly sure whether the ideas presented in the last two posts are entirely correct.

1) I think the Coriolis force does not have anything to do with gravity not being parallel with the centrifugal force. Coriolis force is a "fictional force" which must be introduced in non-inertial, rotating systems for moving objects. Someone wise from [url=http://scienceworld.wolfram.com/physics/CoriolisForce.html]scienceworld.wolfram.com[/url] says it is equal to \[ F = 2m(\vec{v} \times \Omega)  \]

2) The merry-go-round theory would work in case of a projectile or when the air resistance (& gravity) is neglible. However, a balloon strongly depends on the surrounding air, which rotates together with the system (Earth). Well, but the princilple itself should still work in a way: the balloon will not go in a straight line (even if viewed from a non-rotating frame of reference) after it is released from the ground, but its path will curve slightly as it goes up or down (as we know the balloon can control this easily;). I think the best way to look at this is through the conservation of angular momentum.

What do you think?

So kubus ur saying that the air around the ballon when turns with the earth will push the ballon?That's what i thought but then i didn't know what to do after that.....SO Here's the dumb thing i thought,i assumed that the air would not piush it around(Just a new condition to the question) and then the earth rotates under it while it stays put....pretty dumb eh?

I don't think the air is a lot affected by the earth for a smalll period of time,so i don't think the ballon will be pushed in a proper path back to its original position when the earth rotates for a small timeinterval. :huh:

Coriolis force would work even on a planet with no atmosphere. I'm not sure what happens when the balloon goes up or down but when it is hanging still in the air (in the case the air does not affect the motion of the balloon), it should be pulled towards the equator.

To kubus: In our noninertial frame, there is the gravitational force pointing directly towards the center of Earth and the newly introduced Coriolis force caused by the noninertiality of our reference frame - I called this force the centrifugal force in my previous post. This Coriolis force is perpendicular to the axis of Earth's rotation so unless you're on the equator, it is not parallel to the gravitational force. Finally, the balloon produces some lift force which is parallel to the gravitational force. Since the Coriolis force is nonzero, we can't get a zero vector after we add all these three together. I hope you can see how crucial the fact that the Coriolis force in not parallel to the gravitational force is.

To everybody else: I'm sorry if you consider this lengthy explanation a spam :).

Well, if you do not "consider the air to blow it around", then the balloon is really just a very simple rocket.  Gas is being expelled one way, and by the conservation of linear momentum (and hence, Newton's third law), the balloon will simply accelerate in whatever direction its moving in.

As to the spinning of the Earth and the Coriolis effect, I don't know if the balloon would be large enough to experience it significantly.

Now that you say that - we should be able to calculate the Coriolis force quite exactly. I tried to do it and I came to a result

$c=\frac{\omega^2R\sin2\phi}{2}$

where $c$ is the Coriolis acceleration, $\omega$ is angular velocity of Earth's rotation, $R$ the radius of Earth and $phi$ is zero at equator and 90° at north pole. Thus we get:

$c\approx0.0169\times\sin2\phi\text{ m/s}^2$.

Maximum is achieved at 45° and it is about 0.002 g. Not a whole lot...

Bus, I don't exactly get what you mean by your calculation of the coriolis force. There is NO coriolis force when you just stand on the ground, not moving. Or is there? The Coriolis force appears when you actually MOVE in the rotating system, doesn't it?

Confused...

Strangely, thats what I thought too. I have this feeling that I'm calling wrong force the Coriolis force...

OK, no. I got it right but it's for stationary objects only. If you started moving, it would change dramatically. I'll leave it to you to find out why does it work even if you're still (in your rotating system) :). I'd like to see the formula for nonstationary objects though.

Look at my post a bit higher, I gave you the formula and a site where it all should be explained. You can also see last year's FyKoS serial (its beginning) for a bit of explanation (a very terse explanation, as you know FyKoS;). For those who do not know [url=http://fykos.mff.cuni.cz]it's site.[/url]

by the way, this may (or may not) be related, but I think it might be...

I am standing on the ground at some nonequatorial, nonpolar position (so $0 < \phi < \frac{\pi}{2}$). I experience a force $w$ due to my weight towards the center of the Earth and some normal force $N$ directly away from the center. If one considers my motion as a particle on a rotating body, then the net force on me must be $m \cdot R \cos \phi \cdot \omega^2$ directed perpendicular to the axis of rotation (not toward the center of the earth); but how is it possible that the vector sum of the two forces $w$ and $N$, both directed along the line connecting the center of the Earth and me, produce a net force lying along some other line? Is there some other force that I'm overlooking (perhaps even this aforementioned 'Coriolis' force) or what?

[quote="schulmannerism"]Here's the answer I came up with to your question (MysticTerminator), hopefully my argument is correct:
If we were frictionless, we would actually slide down to the equator.
The normal force is only going to push on us just hard enough so that we stay on the surface of the earth--the component perpendicular to the axis of rotation must provide the centripetal acceleration. However, as you observed, the net force $w+N$ is unavoidably radial (on our spherical planet). Therefore there is also a component of the inward net force that is parallel to the axis of rotation. This component has a component pointing towards the equator in the tangential plane. Here are the mathematical details:
The centripetal pseudo-acceleration is
$a_{centrip}=\omega^2 R \cos \phi$
$a_{net} \cos \phi = a_{centrip}$
so
$a_{net} = \omega^2 R$
$a_{parallel} = \omega^2 R \sin \phi$
$a_{tangential} =a_{parallel} \sin \phi = \omega^2 R \sin^2 \phi$[/quote]

so if one considers instead a particle of the lithosphere (for which the situation is effectively identical), does this mean that the Earth is gradually getting more and more flat due to rotation? (of course, I think that $R \omega^2$ is still fairly small, so the sun may engulf it before anything noticeable happens, but theoretically...)

btw, is it just because it's somewhat late for me, or am I getting $a_{\mathrm{net}} = \omega^2 R \cos^2 \phi$, $a_{\mathrm{perpendicular}} = \omega^2 R \cos \phi$, $a_{\mathrm{tangential}} = \omega^2 R \cos \phi \sin \phi$?

No, you got it right MysticTerminator. schulmannerism made a slight mistake: If you take $a_\text{tangential}=a_\text{parallel}\sin\phi$, what do you think happens to the other component of $a_\text{parallel}$? This other component is $a_\text{parallel}\cos\phi$ and it is directed to the center of the Earth - again, you have to divide it into the centripetal component and tangential component, then you'll be left with another, smaller force you haven't taken into account and so on... This is not the right approach, your first mistake was at $a_\text{net}\cos\phi=a_\text{centrip}$. The best thing to do is to write the four equations for the components of $a_\text{net}$ and $a_\text{tangential}$ (that's what I did) or draw a nice picture (and I bet this is what MysticTerminator did :) ):

($\overrightarrow{a}_\text{centripetal}+\overrightarrow{a}_\text{tangential}=\overrightarrow{a}_\text{net}$)

EDIT:
Hm, those colors turned out to look pretty bad after I saved the picture as jpeg, and the way it is scaled doesn't help it either - try clicking it though :).


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $ H$ have index $ 2$ in $ G$ and let $ h\in H$.  Show that $ \#Z_{G}\left(h\right)\equal{}\#Z_{H}\left(h\right)$ or $ \#Z_{G}\left(h\right)\equal{}2\#Z_{H}\left(h\right)$.

I thought about using the Second Isomorphism Theorem, but I couldn't get anywhere.  Does anyone have a hint?  Thank you :)

I think $ Z_{H}(h)$ is centralizer $ h$ in $ H$, or $ C_{H}(h)$  :huh: . Any way for centralizer of $ h$ we have $ C_{H}(h) \equal{} C_{G}(h) \cap H$.


Let ABCD be a fixed convex quadrilateral with BC = DA and BC not parallel with DA. Let
two variable points E and F lie of the sides BC and DA, respectively and satisfy BE = DF.
The lines AC and BD meet at P, the lines BD and EF meet at Q, the lines EF and AC
meet at R.
Prove that the circumcircles of the triangles PQR, as E and F vary, have a common point
other than P.


Let $ABC$ be an isoceles triangle where $AB=AC$. If the angle bisector of $\angle B$ meets $AC$ at $D$ and $BC=BD+AD$ what is the mesure of $\angle A$

[hide]Since  is isoceles and  bisects the vertex angle,  is also the altitude to  and the  is the midpoint of  (proven using congruent triangles).

Therefore,  is a right triangle.  Also, .

By the Pythagorean Thm.,

We know from the problem statement that .  This is equivalent to

From earlier,

So, either BD=0 or CD=0.  A triangle can't have a side of 0, so there is no possible value for

[/hide]

This doesn't seem right.  Did I make a mistake?


Εστω ότι χωρίζουμε το σύνολο Z των ακεραίων σε 2 υποσύνολα $ A,B$, ώστε τα $ A,B$ να μην έχουν κοινά στοιχεία. Να δείξετε ότι τουλάχιστον ένα απ' τα $ A,B$ περιέχουν 2 στοιχεία των οποίων ο μέσος όρος ανήκει στο σύνολο με αυτά.

Υ.Γ.1 Η άσκηση προτείνεται μάλλον για "μικρούς", αλλά πιστεύω θα μπορούσε άνετα να πέσει σε διαγωνισμούς "μεγάλων".
Υ.Γ.2 Αντί για το σύνολο Z, θα μπορούσε να μας δίνεται το σύνολο των φυσικών ή των άρτιων ή και αρκετά άλλα :P

Ας υποθέσουμε ότι ο ισχυρισμός δεν είναι αληθής. Τότε υπάρχουν Α,Β σύνολα ξένα μεταξύ τους ώστε η ένωσή τους να είναι το Ζ και σε κανένα από τα δύο δεν περιέχονται 3 διαδοχικοί όροι αριθμητικής προόδου.
Αν σε ένα από τα Α,Β περιέχονται 2 διαδοχικοί ακέραιοι, αν π.χ. κ,κ+1 είναι στο Α τότε κ-1,κ+2 είναι στο Β οπότε κ-4 είναι στο Α. Εφόσον τώρα κ-4,κ στο Α, αναγκαστικά κ+4 στο Β. Ακόμα κ-4,κ+1 στο Α, άρα κ+6 στο Β. Άρα κ+2,κ+4,κ+6 στο Β, άτοπο.
Αν δύο διαδοχικοί ακέραιοι είναι πάντα σε διαφορετικά σύνολα τότε Α=άρτιοι και Β=περιττοί ή ανάποδα, άτοπο.

Ωραία η λύση σου Παύλο αν και μπορούσε να απλοποιηθείο λίγο αν έλεγες:
Εστω 3 διαδοχικοί άρτιοι. 2 από αυτούς θα ανήκουν στο ίδιο σύνολο (έστω Α) κι έστω ότι αυτοί οι αριθμοί είανι οι $ k, k\plus{}2$ (δεν αλλάζει τίποτα αν είναι $ k,k\plus{}4$). Eτσι, τα $ k\minus{}2,k\plus{}4, k\plus{}1$ θα ανήκουν στο $ B$ (άτοπο)!

Τώρα ας πάμε σε μια πιο δύκολη άσκηση. Εστω ότι χωρίζουμε το σύνολο των μιγαδικών σε 2 σύνολα $ A,B$. Να αποδείξετε ότι υπάρχουν 3 μιγαδικοί, $ z_1,z_2,z_3$, που ανήκουν στο ίδιο σύνολο τέτοιοι ώστε:
$ z_2\minus{}z_1\equal{}i(z_2\minus{}z_3)$!

Μια λυση που μολις βρηκα για το 1ο:

Σε καθε συνολο μπορουμε να βρουμε απειρους αρτιους, διοτι αν οι αρτιοι του ενος συνολου εχουν ενα ανω φραγμα, επιλεγοντας 2 πολυ μεγαλες δυναμεις του 2 του αλλου συνολου και παιρνοντας το μεσο ορο τους καταληγουμε σε ατοπο.
Εστω οτι το Α ειναι το συνολο απο τα 2 που περιεχει το 0, τοτε για α,β αρτιους (α διαφορο του β) στο Β το α+β ειναι επισεις στο Β, διοτι αν δεν ηταν τοτε θα ηταν στο Α το οποιο ειναι ατοπο διοτι ο μεσος ορος του με το 0 εναι το (α+β)/2 που ανηκει στο Α επηδη τα α,β ειναι στο Β. Αρα το α+β ειναι στο Β. Απο αυτο για α,β αρτιους διαφορετικους μεταξυ τους στο Β λαμβανουμε: α + β στο Β, (α + β) + β στο Β, (α + β + β) + α = 2(α+β) στο Β. Ομοια βρησκουμε 3(α+β), 4(α+β) στο Β (τα αθροισματα αυτα ειναι αθροισματα διαφορετικων μεταξυ τους αριθμων του Β, αφου το 0 ανηκει στο Α, ενω μπορουμε χβτγ να παρουμε α,β ομοσημους), το οποιο ειναι ατοπο διοτι το 3(α+β) ειναι ο μεσος ορος των 4(α+β) και 2(α+β).

Μπράβο Πάυλο! Αυτήν την ερμηνεία είχα κι εγώ στο μυαλό μου. ¨Ομως η λύση μου είναι και εδώ λίγο διαφορετική. Βασίζεται στο προηγούμενο ερώτημα.
Βάσει του προηγουμένου λοιπόν στην ευθεία των πραγματικών υπάρχουν 3 σημεία $ A,B,C$ με $ AB \equal{} BC$ που έχουν το ίδιο χρώμα, έστω μπλε. Αν υποθέσουμε ότι η υπόθεση του προβλήματος είναι λάθος, τότε τα σημεία $ D,E,Z$, τέτοια ώστε $ DA \equal{} \perp DB$, $ EB \equal{} \perp EC$, $ ZA \equal{} \perp ZC$ (δηλαδή μέσα των ημικυκλίων με τις αντίστοιχες διαμέτρους), θα είναι και τα 3 κόκκινα. Ομως εύκολα $ DZE$ είναι ορθογώνιο κι ισοσκελές κι άρα φτάσαμε σε άτοπο!

Το πρώτο ερώτημα του bilot4 λέει: Αν οι ακέραιοι χρωματιστουν κόκκινοι/μπλε, τότε υπάρχουν τρεις ακέραιοι με το ίδιο χρώμα οι οποιοι είναι διαδοχικοί όροι αριθμητικής προόδου (με διαφορά =/=0).

Ένα διάσημο θεώρημα του Van Der Waerden λέει: Αν οι ακέραιοι χρωματιστούν με r χρώματα, τότε για κάθε κ, υπάρχουν κ ακέραιοι με το ίδιο χρώμα οι οποίοι είναι διαδοχικοί όροι αριθμητικής προόδου (με διαφορά =/=0).

Το θεώρημα όμως είναι σχετικά δύσκολο για να σας ζητήσω να το αποδείξετε, αλλά καλό είναι να το γνωρίζετε.

[quote="PavlosM"]Ελπίζω να μεταφράζω σωστά και το πρόβλημα να είναι ισοδύναμο με το εξής: Χρωματίζουμε όλα τα σημεία του επιπέδου κόκκινα ή μπλε. Να αποδειχθεί ότι υπάρχει ορθογώνιο και ισοσκελές τρίγωνο με κορυφές του ίδιου χρώματος.
Θεωρούμε τυχαίο τετράγωνο. Αν τρεις κορυφές του είναι του ίδιου χρώματος, τότε ορίζουν ένα τέτοιο τρίγωνο. Αν δύο κορυφές είναι κόκκινες και οι άλλες δύο μπλε, τότε υπάρχουν δύο περιπτώσεις:
Οι δύο κόκκινες κορυφές είναι διαδοχικές. Τότε το κέντρο του τετραγώνου αν είναι κόκκινο, ορίζει με τις κόκκινες κορυφές το ζητούμενο τρίγωνο. Αν πάλι είναι μπλε τότε με τις μπλε κορυφές ορίζει το ζητούμενο τρίγωνο.
Οι δύο κόκκινες κορυφές είναι μη διαδοχικές. Τότε με πλευρά μια διαγώνιο το τετραγώνου κατασκευάζουμε τετράγωνο και εργαζόμαστε όπως την προηγούμενη περίπτωση.[/quote]

Ενα πραμα που το 'χω απορια χρονια τωρα (αλλα ολο ξεχνω να το ρωτησω :P ) ειναι γιατι σε τετοιου ειδους προβληματα ζητειται να αποδειχθει οτι "υπαρχει τριγωνο και οχι "υπαρχουν απειρα τριγωνα". Π.χ. στη συγκεκριμενη ασκηση ο Παυλος αυτο απεδειξε ουσιαστικα, αφου η κατασκευη του γινεται σε πεπερασμενο κομματι του επιπεδου αλλα αφου το επιπεδο ειναι απειρο μπορει να κανει αυτην την κατασκευη απειρες φορες... :?:


sabet konid ke dar har tounomenty ke hich  rasi az darajeye voroody sefr nadarad , 2 ta raase shah darad... :|

lem:har tournment hatman daraye haddeaghal yek rase shah hast.hal an ras ra darnazar migirm manande $x$ hal rase $x$ be tedadi rase khrooji manand$w_{i}$ va az tedadi ras manande $v_{j}$ yal gerefte. midanim baiad:
$d^{-}(v_{i})>0$ hal khode $v_{i}$ ha ra dar nazar begirid beine anha yek shah vojood darad in ras ra $y$ minamim hal be ezaye har $v_{i}$ , $w_{j}$e vojood darad ke $w_{j}\rightarrow\ v_{i}$ pas rase shahi ke dar beine
$v_{i}$ dar nazar migirim in ras be $x$ ba yek yal vasl ast va be har kodam az $w_{j}$ be vasile $x$ vasl ast pas $y$ niz shah hast..


inequalities
function
rearrangement inequality
inequalities proposed

Prove that if $a,b,c$ are positive reals and $abc=1$ then

$\frac{a^3}{\sqrt{7+b^2+c^2}}+\frac{b^3}{\sqrt{7+a^2+c^2}}+\frac{c^3}{\sqrt{7+a^2+b^2}} \ge 1$

Not too elegant, but anyways:

From Holder's inequality, we have

\[\left(\sum_{\text{cyc}} \frac{a^3}{\sqrt{7+b^2+c^2}}\right)^{2/3} \left(\sum_{\text{cyc}}(7+b^2+c^2)\right)^{1/3} \geq a^2+b^2+c^2,\]

and thus it suffices to prove that

\[(a^2+b^2+c^2)^3 \geq \sum_{\text{cyc}} (7+b^2+c^2) = 2(a^2+b^2+c^2)+21.\]

Let $A=a^2$, etc so we need to prove:

\[(A+B+C)^3 \geq 2(A+B+C)+21.\]

From AM-GM,

\begin{eqnarray*}
(A+B+C)^3 &\geq& A^3+B^3+C^3+24 \\
&=& (A^3+1+1)+(B^3+1+1)+(C^3+1+1)+18 \\
&\geq& 3(A+B+C)+18 \\
&\geq& 2(A+B+C)+21
\end{eqnarray*}

Yes,but I prove

$(a+b+c)^3 \ge 21+2(a^2+b^2+c^2)$
by other way!

The best inequality is:

$(a+b+c)^3 \ge \frac{9(a^2+b^2+c^2)}{4} +\frac{81}{4}$

[quote="ThAzN1"]Not too elegant, but anyways:

From Holder's inequality, we have

\[\left(\sum_{\text{cyc}} \frac{a^3}{\sqrt{7+b^2+c^2}}\right)^{2/3} \left(\sum_{\text{cyc}}(7+b^2+c^2)\right)^{1/3} \geq a+b+c,\]

and thus it suffices to prove that

\[(a+b+c)^3 \geq \sum_{\text{cyc}} (7+b^2+c^2) = 2(a^2+b^2+c^2)+21.\]

Now set $x^3=a$, etc so this beomes

$(x^3+y^3+z^3)^3 \geq 2(x^7yz+xy^7z+xyz^7)+21x^3y^3z^3$

$\iff x^9+y^9+z^9+3\sum_{\text{sym}} x^6y^3 \geq 2(x^7yz+xy^7z+xyz^7) +15x^3y^3z^3.$

From AM-GM,

$\sum_{\text{cyc}} (x^9+x^6y^3+x^6z^3) \geq 3\sum_{\text{cyc}} x^7yz \geq 2\sum_{\text{cyc}} x^7yz+3x^3y^3z^3,$ and

$2\sum_{\text{sym}} x^6y^3 \geq 12x^3y^3z^3,$ so the result follows.[/quote]

may i know it is \[\left(\sum_{\text{cyc}} \frac{a^3}{\sqrt{7+b^2+c^2}}\right)^{2/3} \left(\sum_{\text{cyc}}(7+b^2+c^2)\right)^{1/3} \geq a+b+c,\]
or \[\left(\sum_{\text{cyc}} \frac{a^3}{\sqrt{7+b^2+c^2}}\right)^{2/3} \left(\sum_{\text{cyc}}(7+b^2+c^2)\right)^{1/3} \geq a^2+b^2+c^2,\]

[quote="siuhochung"]

may i know it is \[\left(\sum_{\text{cyc}} \frac{a^3}{\sqrt{7+b^2+c^2}}\right)^{2/3} \left(\sum_{\text{cyc}}(7+b^2+c^2)\right)^{1/3} \geq a+b+c,\]
or \[\left(\sum_{\text{cyc}} \frac{a^3}{\sqrt{7+b^2+c^2}}\right)^{2/3} \left(\sum_{\text{cyc}}(7+b^2+c^2)\right)^{1/3} \geq a^2+b^2+c^2,\][/quote]

Ohh, whoops... it's a small mistake but it makes the proof even easier!  :D

Let $A=a^2$, etc so we need to prove:

\[(A+B+C)^3 \geq 2(A+B+C)+21.\]

From AM-GM,

\begin{eqnarray*}
(A+B+C)^3 &\geq& A^3+B^3+C^3+24 \\
&=& (A^3+1+1)+(B^3+1+1)+(C^3+1+1)+18 \\
&\geq& 3(A+B+C)+18 \\
&\geq& 2(A+B+C)+21
\end{eqnarray*}

I've edited the post above.  :)

Ok.Other problem:

$\sum\frac{a^3}{\sqrt{7+b^4+c^4}} \ge 1$

And

$\sum\frac{a}{\sqrt{7+b+c}} \ge 1$

How to solve:

${\sum{a^4(b^3+c^3)^3} \le \frac{8}{81}(a^3+b^3+c^3}^7$

Clear it,please!

Beside,for this inequlity,I have a nice solution quite same as ThAnz1.
 :(  :(  :(
At first,because a bit mistake,I type wrong to
$\sum\frac{a^3}{\sqrt{7+b^2+c^2}} \ge 1$
It's easier.

$(a+b+c)^3 \ge 2(a^2+b^2+c^2)+21$

my proof to the second inequality,
by holder,
${(\sum \frac{a}{\sqrt{7+b+c}})^2 \ge \frac{(a+b+c)^3}{\sum a(7+b+c}}$
it remains to prove
$(a+b+c)^3 \ge 2(ab+bc+ca) + 7(a+b+c)$
Let $x=a+b+c$, we will prove the stronger ineq
$(a+b+c)^3 \ge \frac{2}{3}(a+b+c)^2 + 7(a+b+c)$
equivalently $x(x^2-\frac{2}{3}x-7) \ge 0$
this is true because $x \ge 3$ and $x^2-\frac{2}{3}x-7$ is increasing for $x \ge \frac{1}{3}$, and $3^2-\frac{2}{3}(2)-7=0$, this completes the proof.

To hungkhtn:[tex] (\sum a^4(b^3+c^3))^3\leq\ \frac{8}{81}*(a^3+b^3+c^3)^7[/tex]
I think it is well-known inequality:assume that a^3+b^3+c^3=3,we have [tex]a^4b^3+b^4c^3+c^4a^3\leq\ 3[/tex].See here:
http://www.mathlinks.ro/Forum/topic-28680.html

Ok :) It's true but not really well_known for every one :)

The lastest inequality of mine.I think It's the most difficult of all similar Inequalities I post :

$\frac{a}{\sqrt{7+b^2+c^2}}+\frac{b}{\sqrt{7+a^2+c^2}}+\frac{c}{\sqrt{7+a^2+b^2}} \ge 1$

Please! :)  :)

My solution, applying holder's again in the same way,
\[
\left( {\sum {\frac{a}{{\sqrt {7 + b^2  + c^2 } }}} } \right)^2 \left( {\sum {a(7 + b^2  + c^2 )} } \right) \ge \left( {a + b + c} \right)^3
\]
it remains to prove
\[
\begin{array}{l}
 \left( {a + b + c} \right)^3  \ge \sum\limits_{sym} {a^2 b}  + 7(a + b + c) \\
  \Leftrightarrow \sum {a^3 }  + 2\sum\limits_{sym} {a^2 b}  + 6abc \ge 7(a + b + c) \\
  \Leftrightarrow \sum {a^3 }  + 2\left( {\sum {ab} } \right)\left( {\sum a } \right) \ge 7(a + b + c) \\
 \end{array}
\]
Now,
\[
2\left( {\sum {ab} } \right)\left( {\sum a } \right) \ge 6(a + b + c)
\]
(AMGM)
\[
\sum {a^3 }  \ge 3\left( {\frac{{a + b + c}}{3}} \right)^3  \ge 3\left( {\frac{{a + b + c}}{3}} \right) = a + b + c
\]
(Power mean + AMGM)
Combining them finishes the proof.

Very nice solution.It's simpler than mine alot!
 :)


Let a triangle $ABC$ .We take the points $D,\ E$ such that $BA=BD$ and $CA=CE$ and such that  $D,\ E$ are at the same side with $A$ with respect to $BC$ . Prove that when the area of the quadrilateral $DBCE$ is maximum then
$\angle{DBE}=\angle{DCE}=90$

which quadrilateral are we trying to maximize the area of?


What are the chances that out of a 16 flip sequence of a coin that has equal probability of landing heads or tails that you will never get 2 consecutive heads OR 3 consecutive tails.

Of course (although there are other, perhaps more elegant ways):
[hide="An outline of the method"]For $n \geq 1$, let $a_{n}$ be the number of sequences containing no HH or TTT which end in H, $b_{n}$ be the number of sequences containing no HH or TTT which end in T (but not TT), $c_{n}$ be etc.etc.  The recursions are pretty simple: my $b_{n}$ and $c_{n}$ can be substituted away for very quickly to find $a_{n}$, and it's then computationally simple to calculate $a_{16}$, $b_{16}$, $c_{16}$ and thus the desired probability.  This also opens a window to find the general value when $16$ is replaced by $n$.[/hide]

Some stuff about the general case:

[hide]
Note that there are only two possible ways the sequence can progress:

H (T or TT) H (T or TT) ...

or

(T or TT) H (T or TT) H ...

We can group them like this:

[HT or HTT][HT or HTT][HT or HTT]...[H,HT,HTT] if it starts with heads, and
[TH or TTH][TH or TTH][TH or TTH]... [T, TT, TH,TTH] if it starts with tails.

The number of ways for the first case equals the number of ways to write n-1 as the sum of 2's and 3's (order counts), plus the number of ways to write n-2, plus the number of ways to write n-3.

For the second case, it's the same, except we add the number of ways to write n-2 in twice.

Thus, if $k(N)$ is the number of ways to write $N$ as the sum of 2's and 3's, with order counting, then:

$P(n) = \frac{2k(n-1)+3k(n-2)+2k(n-3)}{2^{n}}$[/hide]

And now a look into the case of 16:

[hide="Solution"]
15=3+2+2+2+2+2+2=3+3+3+2+2+2=3+3+3+3+3
Thus, $k(15)=\binom{7}{1}+\binom{6}{3}+\binom{5}{5}=28$

14=2+2+2+2+2+2+2=3+3+2+2+2+2=3+3+3+3+2
$k(14)=\binom{7}{0}+\binom{6}{2}+\binom{5}{4}=21$

13=3+2+2+2+2+2=3+3+3+2+2
$k(13)=\binom{6}{1}+\binom{5}{3}=16$

$P(16)=\frac{2(28)+3(21)+2(16)}{2^{1}6}=\boxed{\frac{151}{65536}}$

[/hide]

The next step would be using the pattern in the calculations of $k(n)$ above to find an expression for $k$ (not that difficult to do) and finish the generalization.


Hola, me preguntaba, si habia alguien por aqui que haya realizado la olimpiada de mayo de este año, y si sabian cuales habian sido sus puntajes en cada problema.

Los míos fueron:

prob. 1:  10
prob. 2:  10
prob. 3:  3
prob. 4:  0
prob. 5:  5

Como sabes tus puntajes de las preguntas?

Por que como los revisan en cada pais, y mandan algunas de las pruebas y los puntajes, entonces quienes lo revisaron en mi pais, me dijeron despues de mandarlas, cuantos puntos habia tenido en cada problema ^^


geometry
circumcircle
geometric transformation
rotation
ratio
trigonometry
geometry unsolved

This is an old problem, but i want get some ways to solve it. Pleases post your way at here. Thank you very much.

Let $ ABC$ is a triangle. Construct out of $ \triangle ABC$ three equilateral triangles $ BCM, ACN, ABP$. Let $ I, J, K$ are circumcenter of three circles $ (BCM), (CAN), (ABP)$ respectively. Prove that $ IJK$ is an equilateral triangle.

[quote="plane geometry"]see http://mathworld.wolfram.com/NapoleonsTheorem.html[/quote]

Thank you very much, but i don't get any solution at this link.

This is a nice solution for this theorem: http://www.gogeometry.com/Napoleon0.htm

Now i want see more solution. Any body can post a solution at here.

Thank you very much.

[quote="77ant"]my shoddy answer.

$ \triangle{KAJ}\sim\triangle{BAN}$
$ \triangle{KBI}\sim\triangle{ABM}$
$ \triangle{JCI}\sim\triangle{ACM}$

$ \Longrightarrow\frac {KJ}{BN} \equal{} \frac {KI}{AM} \equal{} \frac {JI}{AM}$

$ AM \equal{} BN \equal{} CP$ (well known)
($ \because\triangle{PAC}\sim\triangle{BAN}$, $ \triangle{ACM}\sim\triangle{NCB}$)
$ \therefore\triangle{KJI}$=equilateral[/quote]

Thank you very much. It's a nice solution too.

Who can use the rotation about a point to done this problem?

Oh ,this is a famous theory: like the figure , just prove triangle AIJ,APC,ABN are similar triangle BIK,BPC are similar and triangle CKJ CBN are similar ,with the same ratio $ {\sqrt{3}}$ we get IJ=JK=KI  this theory is called Napoleon Th . that is one of the ways ,we can also use law of cosine to solve it. see solution by shobber at http://www.mathlinks.ro/viewtopic.php?p=339657#339657

If one constructs the equilateral triangles PNX, PMY and MNZ, of centers T, U and V respectively, with NP separating X and M, etc., it is easy to prove:
a) A is the midpoint of MX, similarly for B and C;
b) K is the midpoint of TU, etc.

Also, if D, E and F are the midpoints of NP, MP and NM respectively, then AD = BE = CF ( I posted this property on the forum before, with a very nice and valuable observation of Mr. V, Nicula).


With the trivial observation that $ a_n=2^{2^{n-2}} $ we reduce the inequality to proving that $ (1-\frac{1}{2})(1+\frac{1}{2})...(1+\frac{1}{2^{2^{n-2}}})<1 $ or to $ 1-\frac{1}{2^{2^{n-1}}}<1$ obvious.

Isin't it surprising that a so-easy problem has been given for the BMO?

Pierre.

Well, I do not know about such  a problem given in the Balkan olympiad. Maybe manlio will clarify this.

Maybe because this problem wasn't proposed at BkMO 2003?! Manlio please check your sources more carefully.

The questions from BkMO 2003 are posted on the site, and this problem is not one of them. :)

I found it in the official site of Polish Mathematical Olympiads  http://www.om.edu.pl .
It was one of the problems of the competition called

"XIV Zawody Matematyczne Panstw Baltyckich-Tresci zadan- 2003"

that I have translated as Baltik Mathematical Olympiads or BMO 2003.

Could it be a preolympic competition or an initial step of the official national competition?

Sorry, I missed to say that this competition was made in

Ryga(Lotwa), 2 listopada 2003

Ok, so the problem is the reference you gave in the title ('Balkan')...
For a Baltic competition, this is not a so-big surprise....

Pierre.

manlio please say Baltik MO 2003 rather than BMO. BMO usually stands for British Mathematical Olympiad. BkMO stands for Balkan Mathematical Olympiad. :)

damn, nobody reads that topic with problems sources :?  :)

Sorry, I am so stupid to have  written "Balkan" instead of "Baltic".

Sorry, again for this misprint.


$x,y,z>0$ and $xyz=(1-x)(1-y)(1-z)$
find least value of expression : $A=x^3+y^3+z^3$

it is easy to see should only do when $0<x,y,z<1$
let $\frac{1-x}x=a$,and $b,c$ similar
then $x=\frac 1{a+1}$,$y=\frac 1{b+1}$,$z=\frac 1{c+1}$,
then the problem became $abc=1$
find the minimum of
$\frac 1{(a+1)^3}+\frac 1{(b+1)^3}+\frac 1{(c+1)^3}$,
that is well-known :)

I dont know how well known is that so I did it in the following way:

We have $\frac{xyz}{(1-x)(1-y)(1-z)} = 1$ Hence we can put $\frac{x}{1-x} = \frac {a}{b}$, $\frac{y}{1-y} = \frac {b}{c}$ and $\frac{z}{1-z} = \frac {c}{a}$.

From this we get $x = \frac{a}{a+b}$, $y = \frac{b}{b+c}$ and $z = \frac{c}{a+c}$

So we need to find the minimun of $ \frac{a^{3}}{\left(a+b\right)^{3}}+\frac{b^{3}}{\left(b+c\right)^{3}}+\frac{c^{3}}{\left(c+a\right)^{3}} $ which is $\frac{3}{8}$ according to http://www.artofproblemsolving.com/Forum/viewtopic.php?p=230046&sid=26880e75a0c651e9ebe5ca4eec0d2a7e#p230046


I didn't check but is it different than
http://www.mathlinks.ro/Forum/viewtopic.php?t=19233

Pierre.

It's different!  ;) . Fact as usual we denote $d(n)$ for the number of all divisors of $n$. So when i wrote the problem of Pascual2004 to home, i thought it's my $d(n)$ :D and now this problem is still unsolved to me. !

[quote="pluricomplex"]For each $n$ is a positive integer we denote $d(n)$ is the number of all positive divisors of $n$. Find all $n$ such that there exist a rational number $q$ so that : $d(n)=n^q$.[/quote]
And you think it is possible to solve your problem? ;)

[quote="Pascual2005"]your problem has infinitely many solutions, consider for example numbers of the form $p^{(p-1)}$ with p prime, and lots of other numbers![/quote]
That is why I asked him  :)

[quote]Posted: Sun Dec 26, 2004 4:22 pm    Post subject:

--------------------------------------------------------------------------------

Pascual2005 wrote:
your problem has infinitely many solutions, consider for example numbers of the form  with p prime, and lots of other numbers!

That is why I asked him   :)
[/quote]

I know it too  :D and i have no related material  :( . So i think we could.... ;)


Given a convex cyclic quadrilateral $ ABCD$ with $ AB \equal{} BC \equal{} 1,\ CD \equal{} 2, \ DA \equal{} x\ (0 < x < 4).$ Find the condition for $ x$ such that the center of the excircle of the quadrilateral exists outside the quadrilateral.

I don’t understand well the problem dear [b][size=100]Kunny[/size][/b].

Do you mean the quadrilateral $ ABCD,$ as a bicentric one, with a circumcircle and an excircle? If yes, then its excenter lies always outwardly to it.

In this case, as well known, we have $ AD \minus{} BC \equal{} CD \minus{} AB \equal{} 1$ and so, $ AD \equal{} x \equal{} 2.$

Also, because of $ AD \plus{} BC \equal{} CD \plus{} AB \equal{} 3,$ $ ($ because of $ AD \equal{} 2$ $ ),$ we conclude that the cyclic quadrilateral $ ABCD$ is orthodiagonal , with an incircle and an excircle, as well.

Sorry if I understood something wrong, Kostas Vittas.


$p$ is a prime number prove that $\sum^p_{j=0}\left(\begin {array}{c}p\\j\end{array}\right)\left(\begin{array}{c}p+j\\j\end{array}\right)\equiv 2^p+1(mod p^2)$

Take a look here : [url=http://www.kalva.demon.co.uk/putnam/psoln/psol9110.html]http://www.kalva.demon.co.uk/putnam/psoln/psol9110.html[/url]

But I can't get into the link :( ,maybe my computer doesn't work. :(

There is an http:// to much in it ;-)
try http://www.kalva.demon.co.uk/putnam/psoln/psol9110.html

I have already known it,anyway thank you.Maybe the website is forbidden in China. :D

maybe you should know this:
 for$ P $ as a prime number and $ 0 \leq\ b,d < P $ we have
$\binom{Pa+b}{Pc+d}\equiv$ $\binom{a}{c} \binom{b}{d}\ (mod P)$

yes and thank you. but it is an easy double counting.

You can prove Lucas' theorem out of this one!


superior algebra
superior algebra solved

Hello all - here's a problem that I just can't get started on, though I think that it's probably basic.

Define $Z[\sqrt 2]$ as the set of all reals of the form $a + b\sqrt 2$, where $a, b \in Z$.  Let $\alpha$ be defined such that $\alpha \in Z[\sqrt 2]$, with $\alpha = c + d\sqrt 2$, with $c, d \in Z$.  Define $N[\alpha]$ such that \[ N[\alpha] = \alpha\overline{\alpha} = c^2 - 2d^2. \]

Assume now that $N[\alpha] = n$, where $n \in Z^+$.  Prove that there exists a $\beta \in Z[\sqrt 2]$ such that $N[\beta] = -n$.  Prove also that there exists a $\gamma \in Z[\sqrt 2]$ such that $N[\gamma] = 2n$.

Take $\gamma=(c+2d)+(c+d)\sqrt2$, then $N[\gamma]=-n$.For the
other case, $\gamma=(10c+14d)+(7c+10d)\sqrt2$, then $N[\gamma]=2n$.

Thanks!  Alternately, fix $\gamma = (2c + 2d) + (c + 2d)\sqrt 2$; then, $N[\gamma] = 2n$.

Out of curiosity, what methods did you use to find those answers?

The norm is multiplicative. Since $N(1+\sqrt 2)=-1$ and $N(\sqrt 2)=-2$ (thus $N(2+\sqrt 2)=2$)  we get numbers that work.


Writing a = y + z, b = z + x, c = x + y where x, y, z > 0, the problem reduces to
2\sumsyma + 3\sumsymab \geq (y + z)(y - z) + (z + x)(z - x) + 3(x + y)(y + z)(z + x)
or
2\sumsymx \geq (y + z)(y - z) + (z + x)(z - x) + 6xyz
or
2(x + y + z)  \geq y + z - yz - yz + z + x - xz - xz + 6xyz
or
x + y + yz + yz + xz + xz \geq 6xyz
which is true by AM - GM.

We must prove a^3 + b^3 + c^3 + abc \geq ab^2 + ac^2 + bc^2 + ba^2.

We need only to consider two cases:

1. (c - a)(c - b) \leq 0.
2. (c - a)(c - b) \geq 0.

For 1, we have (a - c)(c - b) \geq 0, which implies

ac + bc \geq c^2 + ab implies

abc + b^2c \geq c^2b + ab^2.

So we need to prove

a^3 + b^3 + c^3 + abc \geq ac^2 + b^2c + ba^2 + abc
<=>
a^3 + b^3 + c^3 \geq a^2b + b^2c + c^2a, which is true by rearrangement inequality.

For 2, we have a^3 + b^3 \geq ab^2 + a^2b by rearrangement.

So we need only to prove c^3 + abc \geq ac^2 + bc^2.

Divide by c: c^2 + ab \geq ac + bc

<=> (c - a)(c - b) \geq 0, which is true by assumption.

In both cases, equality holds iff a = b = c.

haha Arne, I somehow knew someone else was typing up a solution when I was

By Schur, $a^3 + b^3 + c^3 + 3abc \ge a^2b + ab^2 + a^2c + ac^2 + b^2c + bc^2$. By AM-GM, $ -2abc \ge -a^2c - b^2c$. Adding yields the desired inequality.

[quote="Arne"]Writing a = y + z, b = z + x, c = x + y where x, y, z > 0, the problem reduces to
2\sumsyma? + 3\sumsyma?b \geq (y + z)(y - z)? + (z + x)(z - x)? + 3(x + y)(y + z)(z + x)
or
2\sumsymx? \geq (y + z)(y - z)? + (z + x)(z - x)? + 6xyz
or
2(x? + y? + z?)  \geq y? + z? - y?z - yz? + z? + x? - xz? - x?z + 6xyz
or
x? + y? + y?z + yz? + x?z + xz? \geq 6xyz
which is true by AM - GM.[/quote]
rewriting Arne's elegant solution

Writing $a=y+z,b=z+x,c=x+y$ where $x,y,z>0$, the problem reduces to
$2\sum_{sym}a^3+3\sum_{sym}a^2b\geq(y+z)(y-z)^2+(z+x)(z-x)^2+3(x+y)(y+z)(z+x)$
or
$2\sum_{sym}x^3\geq(y+z)(y-z)^2+(z+x)(z-x)^2+6xyz$
or
$2(x^3+y^3+z^3)\geq y^3+z^3-y^2z-yz^2+z^3+x^3-xz^2-x^2z+6xyz$
or
$x^3+y^3+y^2z+yz^2+x^2z+xz^2\geq 6xyz$
which is true by AM - GM

but the point is (as far as I can tell), you don't need the condition that a, b, c are the sides of the triangle

no, it is the necessary condition. :blush:

[quote="Lagrangia"]If a,b,c are the sides of a triangle then:

a^3+b^3+c^3>=a(b-c)^2+b(c-a)^2+3abc.

cheers!

[i]unknown source[/i][/quote]

This is trivially true by Rearrangement, I think  :huh:


Does anyone have any useful programs on the TI-Graphing calculators for mathcounts?

What kinds are you looking for? I mean I have like Pythag. Theo., ya know....

this might be a little off-topic, but can someone help me either tomorrow night or this weekend with hooking up my ti-89 to the computer?

-jorian

me me me but i am busy tomorrow night....so maybe not me


AMC
AIME
geometry
algebra
polynomial
trigonometry
number theory

Here are the results and statistics of Mock AIME 2 according to my answers. The average was a little bit higher, confirming my suspicion that this contest was a little bit easier than Mock AIME 1. The Geometry was but, it looks like I put these problems in about the right order.  I will write a third Mock AIME to be given on some weekend in September. Solutions can be found at http://www.tjhsst.edu/~tmildorf/math/PracticeAIME2Solutions.pdf.

 Thanks to all who participated!

Answers (In spoiler):

[hide]

1. 012

2. 843

3. 437

4. 604

5. 255

6. 351

7. 000 or 400

8. 704

9. 280

10. 727

11. 167

12. 589

13. 071

14. 152

15. 529[/hide]

Participants:

Teddy

super008

JGeneson

t3hg0suazn

Nosepuppy

Optimosis

pi^2

Ultimatemathgeek

joml88

ginasboys

mlee06

interesting_move

theone853

beta

takuanitromars36

Pork_Chop8

JSRosen3

Rep123max

jeffrey

Statistics:

19 People sent answers.

Easiest Problem: Tie: 1, 2. (16 Solvers each)

Hardest Problem: Tie: 10, 14. (Both unsolved)

Easiest Subject: Number Theory (31 Points)

Hardest Subject: Geometry (6 Points)

Mean Score: 4.210526 [hide]= 80/19[/hide]

Mode Score: 5

Median Score: 4

Leaderboard (Those with scores above the average):

super008 9

Pork_Chop8 8

JGeneson 7

Nosepuppy, Teddy 6

JSRosen, Rep123max, interesting_move, Optimosis 5

Score Distribution:

15.

14.

13.

12.

11.

10.

9. *

8. *

7. *

6. **

5. ****

4. **

3. ***

2. ***

1. *

0. *

Solvers Problem by Problem:

1. (E/N) ****************

2. (E/A) ****************

3. (E/C) ***********

4. (M/N) ***

5. (M/N) *******

6. (M/G) ****

7. (M/C) ********

8. (M/N) *****

9. (M/A) ***

10. (H/G)

11. (H/A) ****

12. (H/G) *

13. (H/A) *

14. (H/C)

15. (H/G) *

Comments:  It was very hard for me.
Answers: Sorry, I don't have one.

That was easier than the first one. I am ashamed of myself, I got a 3 again.

The BIGGEST mistake I made was on #13 that I spent so much time on and didn't realize that the polynomial was 11th degree.

Another big mistake I made when I took the test was not trying 4 and 11, both problems I got in like 5 minutes today.

I did better than the first one!  :)  I got a 4 instead of an abysmal 1.  It's still not where I would like it to be but it is a great improvement.  Did anyone else find number 7 kind of easy for a number 7?

yea, #7 wasn't too conceptually difficult.

however, lots of arithematic.  thankfully, i took my time dividing.

I didn't think #7 was number crunchy.  All you need is that 26 choose 13 is divisible by 50 ;-) I think that an answer of 000 probably intimidated some people.

[quote="Peter VDD"]are the results this low? :? I didn't hand it in cuz I only found like 10-12 questions :?[/quote]

Based on the polls,  these Mock AIMEs are considerably harder than the real AIMEs. A score of 10 is excellent.

[quote="Teddy"]for #7, when you say 2 suits are split between two people, can one person have one whole suit and the other have the other whole suit?[/quote]

Yes.  The two suits are completely in their hands right? That's what I was trying to get at, although I guess the word "split" might change the meaning to some folks. I think the answer obtained in the interpretation where the entire suits aren't in one hand each should be accepted. Of course, the other two suits are still in Anders's and Aaron's hands. In this case, $N$ is ${4 \choose 2} \cdot {26 \choose 13} \cdot \left({26 \choose 13} - 1\right) = 649034819756400$ so I will accept the answer $400$ as well as $000$.

In #1, what if we thought you were asking find the largest integer k such that 2004^k divides 2004, and you had the ! to emphasize the importance of the question?

for #15, the solution has a little too much trig to make it too easy to mess up.

i found that [tex]AD [b*cos B + c*cos C] = 2[ABC][/tex] and similar equations.
so changing and adding them gives [tex]\frac {1}{AD} + \frac {1}{BE} + \frac {1}{CF} = \frac {a*cosA+b*cosB+c*cosC} {[ABC]} = \frac {2[ABC]/R} {[ABC]} = \frac {2}R[/tex]
and R= 40*37*13/(4*240) so 2/R=48/481
529

I got 1.  I cannot bear this disgrace. \downto

hmmm?

the packaged solution is pretty clear, imo.

i don't quite understand what you mean, so clarify for me please   ::lol:

[quote="alpha123"]I was just going through the mock aimes and I found this one. I want to see the solutions for this test but the solutions link doesnt seem to be working. Please help![/quote]

You might try [url]http://web.mit.edu/~tmildorf/www/MyMockAIMESeries.pdf[/url]. That particular Mock AIME is covered in pages 16 to 27.

I found it!!!!!

http://wenku.baidu.com/view/5496b04f767f5acfa1c7cd87.html


I got this problem but i don't get how to do it.  It seems that the J should be I which is the $2.6*10^{-10}$ divided by the area of the cross section, but in that case the website won't accept it.  Where am i going wrong?

A small but measurable current of $2.6 * 10^{-10} A$ exists in a copper wire whose diameter is 3.0 mm. The number of charge carriers per unit volume is $8.49 * 10^{28} m^{-3}$. Assuming the current is uniform, calculate the (a) current density and (b) electron drift speed.

nevermind, caught my mistake, forgot it was diameter, not radius


integration
analytic geometry
calculus
trigonometry
geometry
3D geometry
sphere

What is the mean distance between two random points in a unit square?

@ mods: if this is in the wrong forum, please move.

[quote="i_like_pie"]@ mods: if this is in the wrong forum, please move.[/quote]
This seems like as good a place as any - it's definitely calculus. I would assume that "random points in a unit square" refers to the uniform distribution.

Well I know it's

$\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\sqrt{(x_{1}-x_{0})^{2}+(y_{1}-y_{0})^{2}}dx_{0}dx_{1}dy_{0}dy_{1}$

from picking random points $(x_{0}, y_{0})$ and $(x_{1}, y_{1})$, but I couldn't figure out how to evaluate it. Information about this can be found at Mathworld:

http://mathworld.wolfram.com/HypercubeLinePicking.html

It seems quite advanced.

[quote="Kent Merryfield"]I would assume that "random points in a unit square" refers to the uniform distribution.[/quote]
Yes, it is uniform distribution.

[quote="paladin8"][url]http://mathworld.wolfram.com/HypercubeLinePicking.html[/url][/quote]
The answer is on the site. :wink:

[quote="i_like_pie"]What is the mean distance between two random points in a unit square?
[/quote]

We can fix one point and consider mean distance between random point and (for instance) North Pole. Using spherical coordinates we reduce integral for mean to following expression \[\frac{1}{4\pi}\cdot 2\pi \int_{-\pi/2}^{\pi/2}\sqrt{2-2\sin \psi}\cos \psi \, d\psi = \frac{4}{3}.\] I miscalculate possibly.

The Mathworld site states among other things that the mean distance two uniformly picked points in the n-dim unit cube is $O(\sqrt{n})$. It would be interesting to know something about the standard deviation, which is probably $o(\sqrt{n})$ (curse of dimensionality).

Does anybody have any pointers?

[quote="Kent Merryfield"]Alexander - your answer seems to be for a problem involving the unit [b]sphere[/b]. The problem posed here was for the unit [b]square[/b] - and is a considerably harder problem. (Including that you can't do "WLOG let the first point be some specific point".)[/quote]

Thank you, Kent. It's careless mistake.

[quote="hpe"]
The Mathworld site states among other things that the mean distance two uniformly picked points in the n-dim unit cube is O(\sqrt{n}).
[/quote]

I can explain the asymptotoc estimate $\prec \frac{2}{3}\sqrt{n}$ if somebody have interest.

For the original problem, I am pretty close to the answer, but I have left off the last step.

Using paladin8's notation, the random variables $x_{0}$, $x_{1}$, $y_{0}$, and $y_{1}$ have uniform distribution on $[0, 1]$.  Thus $x_{1}-x_{0}$ and $y_{1}-y_{0}$ have the triangular density function $1-|x|$ on $[-1, 1]$.  Thus $(x_{1}-x_{0})^{2}$ and $(y_{1}-y_{0})^{2}$ have the density function $\frac{1}{\sqrt{x}}-1$ on $[0, 1]$.

By convolving that density function with itself, we find that $(x_{1}-x_{0})^{2}+(y_{1}-y_{0})^{2}$ has density $x-4\sqrt{x}+\pi$ on $[0, 1]$ and density $2 \sin^{-1}\frac{2-x}{x}-x+4 \sqrt{x-1}-2$ on $[1, 2]$.

From there it follows that $\sqrt{(x_{1}-x_{0})^{2}+(y_{1}-y_{0})^{2}}$ has density $2x^{3}-8x^{2}+2\pi x$ on $[0, 1]$ and density $4x \sin^{-1}\frac{2-x^{2}}{x^{2}}-2x^{3}+8x\sqrt{x^{2}-1}-4x$ on $[1, \sqrt{2}]$.

That density function seems to agree with the MathWorld article on [url=http://mathworld.wolfram.com/SquareLinePicking.html]Square Line Picking[/url].

The final step is to find the mean of that density function.  The portion on $[0, 1]$ is easy to integrate, but the portion on $[1, \sqrt{2}]$ involves integrating a $\sin^{-1}$ term, which I didn't feel like doing.  If anyone else would like to finish it off, be my guest.


One leg of a right triangle is a diameter of a circle.  Construct a tangent at the point of intersection of the hypotenuse with the circle. Prove that the tangent at the point of intersection of the hypotenuse and the circle bisects the other leg of the triangle.

show that $\overline{AD}=\overline{BD}$ by angle chasing (notice that $\triangle ACE \sim \triangle CDE$)

then $\overline{BC}=\overline{BD}$ (why?)

then the result follows

Join BF.

Then m(BFC) = m(BDC) (since BDFC is a cyclic quadrilateral (why ?))

Also since BD is tangent ot the circle,

         m(BDC) = m(DEC)

        so m(BFC) = m(DEC)

          so BF is parallel to AE and intersects CE at its midpoint F.

         So by midpoint theorem, BF intersects AC at its midpoint.

           In other words BD bisects AC


Set A: Integers from 1 to 100
Set B: Squares from $1^2$ to $100^2$
x is the sum of a pair (a,b) where a is an element of A, b is an element of B.
Set C: Every possible x (for a total of 10000 elements).

What is the mode of C?

(Hope that made sense)

Actually, C have only 7599 elements because some of the elements are the same. From $1^2+1, 1^2+2, ..., 1^2+100, 2^2+2, 2^2+3, ...$ to $50^2+99, 50^2+100$ you have only 2599 different numbers. From $51^2+1$ to $100^2+100$ you have 50x100=5000 different numbers. So C have 7599 numbers. I find out that the first 2599 numbers is the natural numbers from 2 to 2599. Then the next numbers is 2602,..., 2701,2705,...,2804,.......,9901,10001,...,10100.  :)

I meant C to include those duplicates, so it would have 10000 elements. Otherwise it wouldn't have a mode.

You really want C to be a multiset, not a set.  (Just to give you a word you can use in this circumstance.)

An alternative phrasing of the question:

[quote]What number can be written in the form $x + y^2$, where $x$ and $y$ are positive integers no larger than 100, in the largest number of ways?[/quote]

[hide="Hint 1"]Try replacing "100" with smaller numbers.[/hide]
[hide="Hint 2"]You probably want to look at 100 replaced by 1, 4 or 9.[/hide]
[hide="Solution"]Following the hint, we get the idea that the answer is probably 101.  We can write
$101 = 1+10^2 = 20 + 9^2 = 37 + 8^2 = 52 + 7^2 = 65 + 6^2 = 76 + 5^2 = 85 + 4^2 = 92 + 3^2 = 97 + 2^2 = 100 + 1^2$
for a total of 10 different ways.  Suppose there were some number $n$ which could be expressed in 11 different ways.  Then $n = x_i + y_i^2$ for $i = 1, 2, \ldots, 11$.  Now, without loss of generality we have $y_1 > y_2 > \ldots > y_{11}$, and since each is a positive integer, $y_1 > y_{11} + 10$.  But then $x_{11} + y_{11}^2 = x_1 + y_1^2 > x_1 + y_{11}^2 + 20y_{11} + 100 > x_1 + y_{11}^2 + 100$ and so $x_{11} > x_1 + 100$, clearly a contradiction.  So 101 is definitely [i]a[/i] mode of the set.  Following the same argument through with 11 replaced by 10 will leave us in the final stage not with a contradiction but with the unique solution $a_1 = 10, a_{10} = 1$ and so $n = 101$.[/hide]


Fie ABC un triunghi si I un punct in planul lui. O dreapta prin I taie cercurile BIC, CIA, AIB in L,M,N. A', B',C' 3 punte coliniare pe laturile tr ABC si A'', B'', C'' intersectiile dreptelor LA',MB',NC' respectiv cu cercurile BIC, CIA, AIB. Demonstrati ca tangentele in A", B",C" respectiv la BIC, CIA, AIB taie laturile BC, CA, AB in 3 puncte coliniare. :)
             Chiar daca pare o problema "nashpa", aparentele sunt inselatoare :P

Chiar nimeni? :(  Problema e frumoasa sa stiti :wink:  Sa va dau un indiciu:
               Tangenta la cercul circumscris in A, AB, AC si simediana din a (parca) formeaza un fascicol armonic...apar niste rapoarte si cu putina rabdare poate gaseste cineva o solutie :P


Find all the functions $f: R^+\rightarrow R^+(positive\ numbers)$,  such that

$\forall x\in R^+,\ f(f(x))+f^2(x)=12x^4$

 :cool: See you ... :cool:

you can also solve this by sequences... ;)

[quote="Ditdo"]you can also solve this by sequences... [/quote]
Really man ?!:P   What's your solutions ?
 :cool:


I am sure I am missing something obvious. But I cannot find anywhere where it says when Putnam scores are mailed. Does anybody know?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=136975]This topic[/url] was posted on March 22, 2007 - the day I received the 2006 scores in the mail.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=81610]This topic[/url] was posted on March 30, 2006 - the day I received the 2005 scores in the mail.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=30515]This topic[/url] was posted on March 18, 2005 - the day I received the 2004 scores in the mail.

There was no advance notice - just an envelope in my mailbox. So assume the answer to your question is "late March."

Nothing will be mailed directly to any of the participants. The mailing I'm talking about goes to the Putnam supervisor at each college. I get the exact scores (but not a problem-by-problem breakdown) for those who took it at my college, and I get a printout that shows the names and schools of the top ~500 students overall. This list is stratified, but the only thing I'll know is which group each person belonged to, not where he or she ranked within that group. The groupings:

Putnam fellows: the top ~5 students.
N1: the next ~10 students.
N2: the next ~10 students.
Honorable mention: those ranking between about 25 and about 70 or 80.
The rest of the top 100.
The rest of the top 200.
The rest of the top ~500.

Why do they ask for my address if they are not sending me a score?

That printout I referred to of the top 500 students, which gets sent to Putnam supervisors, has addresses on it. I assume it also goes to some graduate school advisers and to some other organizations. I can recall jmerry getting a solicitation to apply for an internship at a company that consulted for the NSA; that was probably triggered by a Putnam list.

My guess is that if you're not on that top 500 list, then your address doesn't matter at all.

I remember getting an "offer" to apply for The Budapest Semesters of Mathematics program a year and a half ago. I am sure others who too were in the top 500 got similar offers.

I'm pretty sure that all of the information on winners and score distributions posted on the AMC website [url=http://www.unl.edu/amc/a-activities/a7-problems/putnamindex.shtml]here[/url] is simply transcribed from the mailing I've described here, shortly after it is received at MIT (which is where Kiran Kedlaya is.)

[quote="Kent Merryfield"][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=136975]This topic[/url] was posted on March 22, 2007 - the day I received the 2006 scores in the mail.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=81610]This topic[/url] was posted on March 30, 2006 - the day I received the 2005 scores in the mail.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=30515]This topic[/url] was posted on March 18, 2005 - the day I received the 2004 scores in the mail.[/quote]
You can add March 13, 2008 to this list.


algebra
polynomial
difference of squares
special factorizations

$[x^{2}+4x+4]-[y^{2}-8+16]$
$(x+2)^{2}-(y-4)^{2}$ factoring two perfect squares
$(x+2+y-4)( x+2-y+4)$ taking difference of squares
left with:

     $(x+y-2)(x-y+6)$

[hide="solution"]$\text{The first part of the equation looks like}$
$(x+2)^{2}$
$\text{and the second half is}$
$-(y-4)^{2}$
$\text{We factor and get difference of squares: }$
$(x+2)^{2}-(y-4)^{2}$
$\text{We factor: }$
$\boxed{(x+y-2)(x-y+6)}$[/hide]


geometry
ratio
perimeter
LaTeX
similar triangles
area of a triangle
Heron\u0027s formula

Given two similar triangles, the area of the larger triangle is sixteen times the area of the smaller triangle. Find the ratio of the perimeter of the larger triangle to the perimeter of the smaller triangle .

[hide="Solution"]Let $ r$ be the desired ratio. By Heron's Formula, we get that the area of the smaller triangle is $ \sqrt{s(s\minus{}a)(s\minus{}b)(s\minus{}c)}$, whereas the area of the larger triangle is $ r^2\sqrt{s(s\minus{}a)(s\minus{}b)(s\minus{}c)}$. So, $ r^2 \equal{} 16$, so $ \boxed{r\equal{}4}$.

[/hide]

[hide="My way of doing it"]

The ratios of the areas of two similar figures are in the form $ \frac{a^2}{b^2}$ where $ \frac{a}{b}$ is the ratio of the sides.  Since the current ration is $ \frac{16}{1}$, this ratio of the sides is $ \frac41$, and because the perimeter is going to be just 3 times $ a$ or $ b$ (the three will cancel), the ratios of the perimeter is then also $ \frac41$.[/hide]

Use area theorem and herons formula

I could not post a solution because I have problems with latex

In general, know that when all the sides of a polygon are
multiplied by $ k$, a positive real number, that its area
is multiplied by $ k^2$.

[quote="darkdieuguerre"][hide="Solution"]Let $ r$ be the desired ratio. By Heron's Formula, we get that the area of the smaller triangle is $ \sqrt {s(s \minus{} a)(s \minus{} b)(s \minus{} c)}$, whereas the area of the larger triangle is $ r^2\sqrt {s(s \minus{} a)(s \minus{} b)(s \minus{} c)}$. So, $ r^2 \equal{} 16$, so $ \boxed{r \equal{} 4}$.

[/hide][/quote]

You don't need to use Heron's Formula.  Just use the fact that when a triangle's dimensions are multiplied by $ k$, the area is multiplied by $ k^2$.


Assume that the power of $ p(x)$ is $ n$, satisfying that $ p(k)\equal{}\frac{1}{\binom{n\plus{}1}{k}},k\equal{}0,1,2,\cdots ,n$
Find $ p(n\plus{}1)$

Yes, it's right I think. I made a mistake when using the Lagrange threorem. Thank you.


Slove this equation: $ cosx\minus{}cos2x\plus{}cos3x\equal{}\frac{1}{2}$

This problem may give you an idea:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=55965[/url]

Equation: $ \Longleftrightarrow cos\frac{x}{2}(cos x \minus{} cos 2x \plus{} cos 3x)\equal{} \frac{1}{2}cos\frac{x}{2}$

$ \Longleftrightarrow \frac{1}{2}cos\frac{3x}{2}\plus{}\frac{1}{2}cos\frac{x}{2}\minus{}\frac{1}{2}cos\frac{5x}{2} \minus{}\frac{1}{2}cos\frac{3x}{2} \plus{} \frac{1}{2}cos\frac{7x}{2}\plus{}\frac{1}{2}cos\frac{5x}{2}\equal{} \frac{1}{2}cos\frac{x}{2}$

$ \Longleftrightarrow cos\frac{7x}{2}\equal{}0$

$ \Longleftrightarrow \frac{7x}{2}\equal{}\frac{\pi}{2}\plus{}k\pi, k\in Z$

$ \Longleftrightarrow x\equal{}\frac{\pi}{7}\plus{}k\frac{2\pi}{7}, k\in Z$


geometry
circumcircle
geometric transformation
homothety
ratio
geometry proposed

How do you mean, the intersection of $AB$ and the tangent at $A$? That's just $A$ right?

Let O be the center and C' the inverse of C wrt $\omega$. $AM\cap\omega=P'$.
$C'P'\cdot C'M=C'A^2=CC'\cdot C'O$. This implies O,C,P',M are concylic. Therefore $\angle CNM=\pi-\angle COM=\angle CPM$. Thus C,M,N,P' are concyclic. This follows that P=P' and so MP always passes through C', or the point of intersection of the tangents at A and B wrt $\omega$

[quote="levi"][quote="levi"]Let $A,B$ be two fixed points which belong to a fixed circle $w$, the midpoint $C$ of the segment $[AB]$ and the intersection $N$ between the line $AB$ with the tangent (noticed $MM$) at the circle $w$ in the point $M$. I note the second intersection $P$ between the circle $w$ with the circumcircle of the triangle $CMN$. Prove that the line $MP$ passes through a fixed point.[/quote]A more nice conclusion of this problem is:
[b]Ascertain the geometric locus of the point [/b]$L\in AP,\ OL\perp AP.$[/quote]
The circle with diameter OA.

[quote="levi"]What speed have you , my dear Mecrazywong ! I don't make angry, but I want a solution without inversion, pole-polar a.s.o. I am in the last class at gymnasium.
\[\overline {\underline {\left| \ \clubsuit\ \heartsuit\  OLD-AGE\ and\ CHILDHOOD\ step\ hand\ in\ hand\ \heartsuit \ \clubsuit\ \right| }}\][/quote]
From the first problem, we deduce that the locus of P is $\omega$. From $L\in AP,OL\perp AP$ L is the midpoint of AP. Note that a homothety centered at A with ratio 1/2 transforms $\omega$ into the circle with diameter OA. Done.

That is right ! Do you know the Russian teacher V. V. Prasolov's book " Zadacii po planimetrii " ? A great book, with a great didactic style of the author ! If you know the Russian language, I recommend you this book.


[quote="xiaoyan"]Hi,
I was just wondering....
In your NORMAL SCHOOL tests, how fast do you guys finish compared to everybody else?

I have entered the state levels of the NCSSM competitions and MATHCOUNTS before, so think I'm pretty above average in math. But it seems I'm always the last to finish in the regular tests for normal school courses. Everybody seems to finish before me...

Is this normal? Or am I getting stupider?[/quote]

Or you are just the only one who knows the stuff and not guessing on each one.

P.S. are you xiaoyan han?

yeah,
it good to FINALLY have my name used in somebody's post in the AoPs Forums.
(I don't know y, but having my name used in a smart ppl forum boosts my ego)
:wow:

Thanks for the encouragement.

Dude you were in my Duke TIP summer e-studies thing :D

Yes.
I peer edited several of you papers too.
How come you weren't in Academic Writing II this summer?

Oh...
I wanted to do that then decided against it because I also wanted to take an AoPs course and a school credit course.
So, was TIP hard?


Proof:

[hide]

Consider the six centers of the circles, and the angles made by joining each of the centers to P. There must be one angle <= 60 degrees in measure (by Pigeonhole). Let this angle be APB, where A and B are centers of respective circles.

Consider the triangle APB. Since <APB <= 60, either all three sides have the same length (in which case AP = AB so circle A includes point B and circle B includes point A), or AB is NOT the largest side of the triangle.

therefore one of AP, BP is larger. WLOG let this be AP. Then AP > AB, so B is contained in circle A.

[/hide]

[hide]

We will use contradiction.  Assume that no circle contained the center of another circle.  Call the 6 centers of the circles A, B, ... , F, and consider angles APB, BPC, ... , FPB.  By our assumption, each one of these angles must be greater than 60 :^o:.  That implies that the sum of those six angles is greater than 360 :^o:, giving us our desired contradiction.

[/hide]

Nukular, that isn't the pigeonhole principle you're using.  It's an idea which may or may not have a name, but definitely not pigeonhole.

"Given a list of real numbers, there is at least one member of the list greater than or equal to the average and at least one less than or equal to the average."

ive always heard it called generalized pigeonhole... because it acts pretty much the same as far as the proof goes...

According to MathWorld, it's called the Fubini Principle.

Hmm interesting, in my combinatorics book it calls the Fubini principle the pidgeonhole principle. But whatever, names are names it all matters on how we use the underlying principles.

Man, I wish my name was Fubini, Fubu, Fubanini, Fubanunu, Funani.... Behold! The great Fubini!


function
trigonometry
algebra
domain
calculus
derivative
real analysis

I don't think it can be true. If inverse exists then $ det Df(a)$ mustn't be 0, so if I just consider a function where its jacobian at a is nonzero but has zero determinant, your statement is impossible already.

The Local Inverse Function Theorem reads as following: [i]Let $ X,Y$ be Banach spaces, $ U \subset X$  and $ f : X \to Y$ belongs to $ C^1(U;Y)$. If $ f'(a)$ is an isomorphism of $ X$ onto $ Y$ for $ a \in U$ then $ f$ has its inverse.[/i]

Let us consider the following example
\[ f\left( x \right) = \left\{\begin{gathered} x + 2x^2 \sin \frac {1} {x},x \ne 0 \hfill \\
0,x = 0 \hfill \\
\end{gathered} \right.
\]
then if $ U$ is a neighborhood of $ 0$ then $ f'(x) = 0$ has a solution in $ U$ but $ f' '(x) \ne 0$ at such solution. This means that $ f$ is not injective on $ U$. That's all.  :roll:

Dear Soarer and bookworm: We need to have $ f\in C^{1}$, but $ f^{\minus{}1}$ it not necessarily differentiable.

[quote="hungkhtn"]Let $ f: U\subset R^n\to R^n$ such that $ U$ is open and $ f$ is $ C^1$ in $ U$.  $ a$ in $ U$ such that $ Df(a)\not \equal{} 0$ (note, not $ det Df(a)\not \equal{} 0$). Prove that there exists inverse funtion $ f^{ \minus{} 1}: B\to A,$ where $ A$ is an open ball centered at $ a$, and $ B$ is an open ball containing $ f(a)$.[/quote]
Take a small enough ball $ A$ center at $ a$. $ A$ is convex then $ f|_A$ is injective, and by Brower's theorem on domain invariance, then just put $ B\equal{}f(A)$.

But why $ f$ is injective? Any more clear idea, since this is the main part of solution, of course.

This's just a technique part: Any continuos funtion $ g$ on $ [0;1]$ to $ \mathbb R$, with $ f'(t)\neq0$ is injective.

$ f\in C^1$ and $ Df(a)\neq0$, so there's a ball $ A$ center at $ a$ which $ Df(x)\neq0$ for any $ x\in A$. Let $ f\equal{}(f_1,...,f_n)$. For $ x\neq y$ in $ A$ which $ f(x)\equal{}f(y)$ so $ f_i(x)\equal{}f_i(y)$. Consider a segment $ \gamma: [0;1]\to A$ which $ \gamma(t)\equal{}tx\plus{}(1\minus{}t)y$ and then $ f_i\circ\gamma: [0;1] \to \mathbb R$ then ...

[quote="pluricomplex"]This's just a technique part: Any continuos funtion $ g$ on $ [0;1]$ to $ \mathbb R$, with $ f'(t)\neq0$ is injective....[/quote]

for some $ t$ or all $ t$?

Nice idea, plurix. But even when $ Df(a)\not\equal{}0$, $ \gamma'$ can be zero. The derivative is the product of two matrixes (linear functions), not of real numbers.

hmm it's really bad. The problem is false for example $ f\equal{}(x_1,0,...,0)$. With my arguments above, if $ f_{x_i}\neq0$ for all $ i$ then the problem holds.