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let $\triangle ABC$, $\triangle A'B'C'$ be two triangles, and $\omega$, $\omega '$ be their circumcircles.
does there exist a projective transformation that maps $A$ to $A'$, $B$ to $B'$, $C$ to $C'$ and $\omega$ to $\omega '$ ?

i don`t know the answer to this question. maybe someone knows ;)

[quote="maky"]let $\triangle ABC$, $\triangle A'B'C'$ be two triangles, and $\omega$, $\omega '$ be their circumcircles.
does there exist a projective transformation that maps $A$ to $A'$, $B$ to $B'$, $C$ to $C'$ and $\omega$ to $\omega '$ ?[/quote]

Yes, there exists one; note that this is pretty much equivalent to http://www.mathlinks.ro/Forum/viewtopic.php?t=28189 . I am going to post a proof when I find time (should be in about a week).

  darij

that`s good to hear :)
i thought it is somehow equivalent, i posted a "proof" there (the computations aren`t finished and it may be totally mistaken).
anyways, i posted it in this form for things to become clear.
my idea was to consider an "inversion" (i am not sure if i can call it that way) in three dimensions. i take $\omega$ and $\triangle ABC$ inscribed in it, and i take $T$ outside the plane of $\omega$. if i choose a "good" $T$, then the image of $\triangle ABC$ will be some triangle similar to $A'B'C'$, and $\omega$ will map to a circle.
however, finding such a point $T$ was not easy for me.

thanks for the reply

I promised to post a proof, let's keep promises...

Warning: this is going to get VERY BORING, but I have never written up this kind of proof yet, so... try to make out the best of this.

[color=blue][b]Theorem 1.[/b] Let $ K$ be a field where $ \minus{} 1$ is not a square. We work in the Euclidean plane $ K^{2}$.

Let $ ABC$ and $ A_{1}B_{1}C_{1}$ be two non-degenerate triangles, and let $ k$ and $ k_{1}$ be the respective circumcircles of these two triangles. Then, there exists a projective transformation which maps the points $ A$, $ B$, $ C$ to the points $ A_{1}$, $ B_{1}$, $ C_{1}$, respectively, and maps the circle $ k$ to the circle $ k_{1}$.[/color]

[i]Proof of Theorem 1.[/i] First a lemma:

[color=blue][b]Theorem 2.[/b] Let $ K$ be a field. We work in the Euclidean plane $ K^{2}$.

Let $ A$, $ B$, $ C$, $ D$, $ A_{0}$, $ B_{0}$, $ C_{0}$, $ D_{0}$ be eight points such that no three of the points $ A$, $ B$, $ C$, $ D$ are collinear and no three of the points $ A_{0}$, $ B_{0}$, $ C_{0}$, $ D_{0}$ are collinear. Then, there exists a projective transformation which maps the points $ A$, $ B$, $ C$, $ D$ to the points $ A_{0}$, $ B_{0}$, $ C_{0}$, $ D_{0}$, respectively.[/color]

[i]Proof of Theorem 2.[/i] Theorem 2 should appear in any serious book on projective geometry, but since I don't have one at hand right now, here comes my proof:

In the following, we use the abbreviation $ \left[ P_{1}P_{2}P_{3}\right]$ for [b]twice[/b] the directed area of a triangle $ P_{1}P_{2}P_{3}$. I say "twice" because the directed area of a triangle is usually defined as a certain determinant divided by $ 2$, but we cannot divide by $ 2$ if our field $ K$ happens to have characteristic $ 2$, so we don't divide at all and just denote the determinant by $ \left[ P_{1}P_{2}P_{3}\right]$.

Let's work in barycentric coordinates with respect to triangle $ ABC$. This means that we assign to every point $ P\in K^{2}$ the following set:

$ \text{Bar}_{ABC}\left( P\right) \equal{} \left\{\left( \begin{array}[c]{c}\lambda\cdot\left[ BPC\right] \\
\lambda\cdot\left[ CPA\right] \\
\lambda\cdot\left[ APB\right] \end{array}\right) \mid\lambda\in K\setminus\left\{ 0\right\}\right\}$.

This set $ \text{Bar}_{ABC}\left( P\right)$ is a set of $ 3$-dimensional vectors; each of these vectors is called a [i]barycentric coordinate vector[/i] of the point $ P$ with respect to triangle $ ABC$. Therefore, every point $ P\in K^{2}$ has many barycentric coordinate vectors with respect to triangle $ ABC$ (if $K$ is an infinite field, then these are infinitely many vectors); these vectors are all proportional to each other, i. e.: If $ \mathbf{p}_{1}\in\text{Bar}_{ABC}\left( P\right)$ and $ \mathbf{p}_{2}\in\text{Bar}_{ABC}\left( P\right)$, then there exists a nonzero $ \mu\in K$ such that $ \mathbf{p}_{1} \equal{} \mu\mathbf{p}_{2}$. Conversely, if $ \mathbf{p}\in\text{Bar}_{ABC}\left( P\right)$, then, for each nonzero $ \mu\in K$, we have $ \mu\mathbf{p}\in\text{Bar}_{ABC}\left( P\right)$ as well.

For every vector $ \mathbf{v} \equal{} \left( \begin{array}[c]{c}x \\
y \\
z \end{array}\right)$ with $ x \plus{} y \plus{} z\neq0$, there exists one and only one point $ P\in K^{2}$ such that $ \mathbf{v}\in\text{Bar}_{ABC}\left( P\right)$. Vectors $ \mathbf{v} \equal{} \left(\begin{array}[c]{c}x \\
y \\
z \end{array}\right)$ such that $ \mathbf{v}\neq0$ but $ x \plus{} y \plus{} z \equal{} 0$ can be thought of as barycentric coordinate vectors of infinite points, and we won't take particular care of these in the following.

Note that in geometry, the assertion that $ \left(\begin{array}[c]{c}x \\
y \\
z \end{array}\right) \in\text{Bar}_{ABC}\left( P\right)$ for some point $ P$ is usually stated as follows: "The point $ P$ has barycentric coordinates $ \left( x: y: z\right)$ with respect to triangle $ ABC$. " Hereby, $ \left( x: y: z\right)$ is a triple of elements of $ K$ determined up to proportionality by a nonzero factor (for instance, $ \left( 1: 1: 2\right)$ is said to be equal to $ \left( 2: 2: 4\right)$). However, we are going to stick with the notation $ \left( \begin{array}[c]{c}x \\
y \\
z \end{array}\right) \in\text{Bar}_{ABC}\left( P\right)$ since we want to work with vectors.

Since $ A$, $ B$, $ C$ are the vertices of triangle $ ABC$, they have the vectors $ \mathbf{a} \equal{} \left( \begin{array}[c]{c}1 \\
0 \\
0 \end{array}\right)$, $ \mathbf{b} \equal{} \left( \begin{array}[c]{c}0 \\
1 \\
0 \end{array}\right)$, $ \mathbf{c} \equal{} \left( \begin{array}[c]{c}0 \\
0 \\
1 \end{array}\right)$ as their barycentric coordinate vectors with respect to triangle $ ABC$, respectively; i. e., we have $ \mathbf{a}\in\text{Bar}_{ABC}\left( A\right)$, $ \mathbf{b}\in\text{Bar}_{ABC}\left(B\right)$ and $ \mathbf{c}\in\text{Bar}_{ABC}\left( C\right)$. Let $ \mathbf{d}\in\text{Bar}_{ABC}\left( D\right)$, $ \mathbf{a}_{0}\in\text{Bar}_{ABC}\left( A_{0}\right)$, $ \mathbf{b}_{0}\in\text{Bar}_{ABC}\left( B_{0}\right)$, $ \mathbf{c}_{0}\in\text{Bar}_{ABC}\left( C_{0}\right)$, $ \mathbf{d}_{0}\in\text{Bar}_{ABC}\left( D_{0}\right)$ be barycentric coordinate vectors of the points $ D$, $ A_{0}$, $ B_{0}$, $ C_{0}$, $ D_{0}$ with respect to triangle $ ABC$.

For any three $ 3$-dimensional vectors $ \mathbf{u}$, $ \mathbf{v}$, $ \mathbf{w}$, we denote by $ \left( \mathbf{u},\mathbf{v},\mathbf{w}\right)$ the $ 3\times3$ matrix whose first, second and third column-vectors are $ \mathbf{u}$, $ \mathbf{v}$, $ \mathbf{w}$, respectively. Then, it is known that three points $ U$, $ V$, $ W$ with barycentric coordinate vectors $ \mathbf{u}\in\text{Bar}_{ABC}\left( U\right)$, $ \mathbf{v}\in \text{Bar}_{ABC}\left( V\right)$, $ \mathbf{w}\in \text{Bar}_{ABC}\left( W\right)$ are collinear if and only if the matrix $ \left( \mathbf{u},\mathbf{v},\mathbf{w}\right)$ has determinant $ 0$. Since the points $ A_{0}$, $ B_{0}$, $ C_{0}$ are not collinear (since no three of the points $ A_{0}$, $ B_{0}$, $ C_{0}$, $ D_{0}$ are collinear), it thus follows that the matrix $ \left( \mathbf{a}_{0},\mathbf{b}_{0},\mathbf{c}_{0}\right)$ has determinant $ \neq0$. Hence, this matrix is invertible, so there exists a vector $ \left(\begin{array}[c]{c}u \\
v \\
w \end{array}\right) \neq0$ such that $ \left( \mathbf{a}_{0},\mathbf{b}_{0},\mathbf{c}_{0}\right) \left( \begin{array}[c]{c}u \\
v \\
w \end{array}\right) \equal{} \mathbf{d}_{0}$. Now, $ \left( \mathbf{a}_{0},\mathbf{b}_{0},\mathbf{c}_{0}\right) \left(\begin{array}[c]{c}u \\
v \\
w \end{array}\right) \equal{} u\mathbf{a}_{0} \plus{} v\mathbf{b}_{0} \plus{} w\mathbf{c}_{0}$, as can be easily verified by computation. Hence, $ u\mathbf{a}_{0} \plus{} v\mathbf{b}_{0} \plus{} w\mathbf{c}_{0} \equal{} \mathbf{d}_{0}$. Thus, $ u\neq0$, $ v\neq0$ and $ w\neq0$ (in fact, if, for instance, we had $ u \equal{} 0$, then $ u\mathbf{a}_{0} \plus{} v\mathbf{b}_{0} \plus{} w\mathbf{c}_{0} \equal{} \mathbf{d}_{0}$ would become $ v\mathbf{b}_{0} \plus{} w\mathbf{c}_{0} \equal{} \mathbf{d}_{0}$, what would yield $ D_{0}\in B_{0}C_{0}$, but this is impossible since no three of the points $ A_{0}$, $ B_{0}$, $ C_{0}$, $ D_{0}$ are collinear).

Now let's write out the vector $ \mathbf{d}$ in its coordinates: $ \mathbf{d} \equal{} \left(\begin{array}[c]{c}U \\
V \\
W \end{array}\right)$. This rewrites as $ \mathbf{d} \equal{} U\left( \begin{array}[c]{c}1 \\
0 \\
0 \end{array}\right) \plus{} V\left( \begin{array}[c]{c}0 \\
1 \\
0 \end{array}\right) \plus{} W\left( \begin{array}[c]{c}0 \\
0 \\
1 \end{array}\right) \equal{} U\mathbf{a} \plus{} V\mathbf{b} \plus{} W\mathbf{c}$. We have $ U\neq0$, $ V\neq0$ and $ W\neq0$ (in fact, if, for instance, we had $ U \equal{} 0$, then $ U\mathbf{a} \plus{} V\mathbf{b} \plus{} W\mathbf{c} \equal{} \mathbf{d}$ would become $ V\mathbf{b} \plus{} W\mathbf{c} \equal{} \mathbf{d}$, what would yield $ D\in BC$, but this is impossible since no three of the points $ A$, $ B$, $ C$, $ D$ are collinear).

Now, barycentric coordinates (extended to the projective plane $ \mathbb{P}\left( K^{3}\right)$) are a particular case of homogeneous coordinates; hence, as in any system of homogeneous coordinates, projective transformations are represented by $ 3\times3$ matrices in barycentric coordinates (not uniquely represented, but we won't need that). Hence, if we denote by $ \varphi$ the transformation which maps every point $ P$ to the point $ Q$ such that $ \left( \frac {u}{U}\mathbf{a}_{0},\frac {v}{V}\mathbf{b}_{0},\frac {w}{W}\mathbf{c}_{0}\right) \mathbf{p}\in\text{Bar}_{ABC}\left( Q\right)$, where $ \mathbf{p}$ is an element of $ \text{Bar}_{ABC}\left( P\right)$ (this transformation is well-defined: different choices of $ \mathbf{p}$ from the set $ \text{Bar}_{ABC}\left( P\right)$ yield the same $ Q$, as can be easily seen, but sometimes $ Q$ will be an infinite point), then this transformation $ \varphi$ is projective.

Now, $ \mathbf{a}\in\text{Bar}_{ABC}\left( A\right)$, and

$ \left( \frac {u}{U}\mathbf{a}_{0},\frac {v}{V}\mathbf{b}_{0},\frac {w}{W}\mathbf{c}_{0}\right) \mathbf{a} \equal{} \left( \frac {u}{U}\mathbf{a}_{0},\frac {v}{V}\mathbf{b}_{0},\frac {w}{W}\mathbf{c}_{0}\right) \left(\begin{array}[c]{c}1 \\
0 \\
0\end{array}\right) \equal{} \frac {u}{U}\mathbf{a}_{0}\in\text{Bar}_{ABC}\left( A_{0}\right)$

(since $ \mathbf{a}_{0}\in\text{Bar}_{ABC}\left( A_{0}\right)$, and $ \frac {u}{U}\in K$ is nonzero since $ u\neq0$). Thus, $ \varphi\left( A\right) \equal{} A_{0}$. Similarly, $ \varphi\left( B\right) \equal{} B_{0}$ and $ \varphi\left( C\right) \equal{} C_{0}$. Finally, $ \mathbf{d}\in\text{Bar}_{ABC}\left( D\right)$, and

$ \left( \frac {u}{U}\mathbf{a}_{0},\frac {v}{V}\mathbf{b}_{0},\frac {w}{W}\mathbf{c}_{0}\right) \mathbf{d} \equal{} \left( \frac {u}{U}\mathbf{a}_{0},\frac {v}{V}\mathbf{b}_{0},\frac {w}{W}\mathbf{c}_{0}\right) \left(\begin{array}[c]{c}U \\
V \\
W \end{array}\right)$
$ \equal{} U\cdot\frac {u}{U}\mathbf{a}_{0} \plus{} V\cdot\frac {v}{V}\mathbf{b}_{0} \plus{} W\cdot\frac {w}{W}\mathbf{c}_{0} \equal{} u\mathbf{a}_{0} \plus{} v\mathbf{b}_{0} \plus{} w\mathbf{c}_{0} \equal{} \mathbf{d}_{0}\in\text{Bar}_{ABC}\left( D_{0}\right)$.

Thus, $ \varphi\left( D\right) \equal{} D_{0}$. Hence, altogether, the projective transformation $ \varphi$ maps the points $ A$, $ B$, $ C$, $ D$ to the points $ A_{0}$, $ B_{0}$, $ C_{0}$, $ D_{0}$, respectively. This proves Theorem 2.

Now we continue with the proof of Theorem 1. This proof can be done in several ways, mostly requiring some knowledge about conics and some knowledge about triangles. Here is one that requires the minimum of the former but quite an amount of the latter:

Let $ L$ and $ L_{1}$ be the symmedian points of triangles $ ABC$ and $ A_{1}B_{1}C_{1}$. Let the lines $ AL$, $ BL$, $ CL$ meet the circumcircle $ k$ of triangle $ ABC$ at the points $ X^{\prime}$, $ Y^{\prime}$, $ Z^{\prime}$ (different from $ A$, $ B$, $ C$). Let the lines $ A_{1}L_{1}$, $ B_{1}L_{1}$, $ C_{1}L_{1}$ meet the circumcircle $ k_{1}$ of triangle $ A_{1}B_{1}C_{1}$ at the points $ X_{1}^{\prime}$, $ Y_{1}^{\prime}$, $ Z_{1}^{\prime}$ (different from $ A_{1}$, $ B_{1}$, $ C_{1}$).

We notice without proof four facts which follow from the condition that $ \minus{} 1$ is not a square in $ K$ (these facts are all pretty obvious if $ K \equal{} \mathbb{R}$, so chances are high you are not interested in their proofs):

[b](1)[/b] The square of the distance between any two distinct points in $ K^{2}$ is nonzero. (Note that the distance itself is not necessarily defined, but the square of the distance between two points $ P_{1}$ and $ P_{2}$ with Cartesian coordinates $ P_{1}\left( x_{1},y_{1}\right)$ and $ P_{2}\left( x_{2},y_{2}\right)$ is always defined as $ \left( x_{1} \minus{} x_{2}\right)^{2} \plus{} \left( y_{1} \minus{} y_{2}\right)^{2}$.)

[b](2)[/b] Any three non-collinear points in $ K^{2}$ lie on one circle which is non-degenerate (i. e. doesn't split into two lines).

[b](3)[/b] The symmedian point of a non-degenerate triangle in $ K^{2}$ does not lie on any sideline of the triangle.

[b](4)[/b] The five points $ A_{1}$, $ B_{1}$, $ C_{1}$, $ X_{1}^{\prime}$, $ Y_{1}^{\prime}$ are pairwise distinct, and no three of them are collinear.

No three of the points $ A$, $ B$, $ C$, $ L$ are collinear (since triangle $ ABC$ is non-degenerate, and [b](3)[/b] shows that the symmedian point $ L$ of triangle $ ABC$ does not lie on any sideline of the triangle $ ABC$). Similarly, no three of the points $ A_{1}$, $ B_{1}$, $ C_{1}$, $ L_{1}$ are collinear.

Hence, according to Theorem 2, there exists a projective transformation $ p$ which maps the points $ A$, $ B$, $ C$, $ L$ to the points $ A_{1}$, $ B_{1}$, $ C_{1}$, $ L_{1}$, respectively. Thus, $ p\left( A\right) \equal{} A_{1}$, $ p\left( B\right) \equal{} B_{1}$, $ p\left( C\right) \equal{} C_{1}$ and $ p\left( L\right) \equal{} L_{1}$.

Now we are going to prove that $ p\left( k\right) \equal{} k_{1}$.

Since $ p$ is a projective transformation, it maps lines to lines. Therefore, since $ p\left( A\right) \equal{} A_{1}$, $ p\left( B\right) \equal{} B_{1}$, $ p\left( C\right) \equal{} C_{1}$ and $ p\left( L\right) \equal{} L_{1}$, we must also have $ p\left( AL\cap BC\right) \equal{} A_{1}L_{1}\cap B_{1}C_{1}.$ Denoting $ D^{\prime} \equal{} AL\cap BC$ and $ D_{1}^{\prime} \equal{} A_{1}L_{1}\cap B_{1}C_{1}$, this becomes $ p\left( D^{\prime}\right) \equal{} D_{1}^{\prime}$.

Applying Theorem 10 from [1] to triangle $ ABC$, we get $ \frac {AL}{LD^{\prime }}: \frac {AX^{\prime}}{X^{\prime}D^{\prime}} \equal{} \minus{} \frac {1}{2}$ (with directed lengths). Applying the same theorem to triangle $ A_{1}B_{1}C_{1}$, we get $ \frac {A_{1}L_{1}}{L_{1}D_{1}^{\prime}}: \frac {A_{1}X_{1}^{\prime}}{X_{1}^{\prime}D_{1}^{\prime}} \equal{} \minus{} \frac {1}{2}$. Thus,

[color=green][b](5)[/b][/color] $ \frac {AL}{LD^{\prime}}: \frac {AX^{\prime}}{X^{\prime}D^{\prime}} \equal{} \frac {A_{1}L_{1}}{L_{1}D_{1}^{\prime}}: \frac {A_{1}X_{1}^{\prime}}{X_{1}^{\prime}D_{1}^{\prime}}$.

Now, let $ S \equal{} p\left( X^{\prime}\right)$. We know that the transformation $ p$ is projective, and thus it maps lines to lines. Hence, since $ p\left( A\right) \equal{} A_{1}$, $ p\left( L\right) \equal{} L_{1}$, $ p\left( D^{\prime}\right) \equal{} D_{1}^{\prime}$ and $ p\left( X^{\prime}\right) \equal{} S$, the relation $ X^{\prime}\in AL$ yields $ S\in A_{1}L_{1}$, and since cross-ratios are invariant under projective transformations, we also get $ \frac {A_{1}L_{1}}{L_{1}D_{1}^{\prime}}: \frac {A_{1}S}{SD_{1}^{\prime}} \equal{} \frac {AL}{LD^{\prime}}: \frac {AX^{\prime}}{X^{\prime}D^{\prime}}$. Combining this with [color=green][b](5)[/b][/color], we get $ \frac {A_{1}L_{1}}{L_{1}D_{1}^{\prime}}: \frac {A_{1}S}{SD_{1}^{\prime}} \equal{} \frac {A_{1}L_{1}}{L_{1}D_{1}^{\prime}}: \frac {A_{1}X_{1}^{\prime}}{X_{1}^{\prime}D_{1}^{\prime}}$, so that $ \frac {A_{1}S}{SD_{1}^{\prime}} \equal{} \frac {A_{1}X_{1}^{\prime}}{X_{1}^{\prime}D_{1}^{\prime}}$. Since both points $ S$ and $ X_{1}^{\prime}$ lie on the line $ A_{1}D_{1}^{\prime}$, and the lengths are directed, this equation yields $ S \equal{} X_{1}^{\prime}$. Since $ S \equal{} p\left( X^{\prime}\right)$, this becomes $ X_{1}^{\prime} \equal{} p\left( X^{\prime}\right)$. Similarly, $ Y_{1}^{\prime} \equal{} p\left( Y^{\prime}\right)$ and $ Z_{1}^{\prime} \equal{} p\left( Z^{\prime}\right)$.

The five points $ A_{1}$, $ B_{1}$, $ C_{1}$, $ X_{1}^{\prime}$, $ Y_{1}^{\prime}$ lie on the circumcircle $ k_{1}$ of triangle $ A_{1}B_{1}C_{1}$. In other words, the circle $ k_{1}$ is a conic passing through these five points. Since these five points are pairwise distinct and no three of them are collinear (from [b](4)[/b]), there exists exactly one conic passing through these five points. Hence, the circle $ k_{1}$ is the only conic passing through the five points $ A_{1}$, $ B_{1}$, $ C_{1}$, $ X_{1}^{\prime}$, $ Y_{1}^{\prime}$. Since $ A_{1} \equal{} p\left( A\right)$, $ B_{1} \equal{} p\left( B\right)$, $ C_{1} \equal{} p\left( C\right)$, $ X_{1}^{\prime} \equal{} p\left( X^{\prime}\right)$, $ Y_{1}^{\prime} \equal{} p\left( Y^{\prime}\right)$, this becomes: The circle $ k_{1}$ is the only conic passing through the five points $ p\left( A\right)$, $ p\left( B\right)$, $ p\left( C\right)$, $ p\left( X^{\prime}\right)$, $ p\left( Y^{\prime}\right)$.

Now, the five points $ A$, $ B$, $ C$, $ X^{\prime}$, $ Y^{\prime}$ lie on the circumcircle $ k$ of triangle $ ABC$. Thus, the circle $ k$ is a conic passing through the five points $ A$, $ B$, $ C$, $ X^{\prime}$, $ Y^{\prime}$. Hence, since the transformation $ p$ is projective and therefore maps conics to conics, the image of the circle $ k$ under the transformation $ p$ must be a conic passing through the five points $ p\left( A\right)$, $ p\left( B\right)$, $ p\left( C\right)$, $ p\left( X^{\prime}\right)$, $ p\left( Y^{\prime}\right)$. But the only conic passing through these five points $ p\left( A\right)$, $ p\left( B\right)$, $ p\left( C\right)$, $ p\left( X^{\prime}\right)$, $ p\left( Y^{\prime}\right)$ is the circle $ k_{1}$. Hence, the image of the circle $ k$ under the transformation $ p$ is the circle $ k_{1}$.

Altogether, our projective transformation $ p$ maps the points $ A$, $ B$, $ C$ to the points $ A_{1}$, $ B_{1}$, $ C_{1}$, respectively, and maps the circle $ k$ to the circle $ k_{1}$. This proves Theorem 1.

[b]References[/b]

[1] Darij Grinberg, [i]Three properties of the symmedian point[/i].
http://www.cip.ifi.lmu.de/~grinberg/geometry2.html

  Darij

wow. thanks for the proof. i think i understood most of it... (i am not very familiar with all the things there)
i knew the lemma, but i thought that it can be proved "geometrically".
the result is very nice indeed.
also, i thought at something "easier" when the "new" triangle is equilateral.
let $ABC$ be the triangle and $\omega$ the circumcircle. also, let the tangents in $A,B,C$ at $\omega$ meet the sides $BC,CA,AB$ in $M,N,P$. then $M,N,P$ are collinear. take the projective transformation that keeps $\omega$ fixed and maps line $MN$ to the infinite line (i`m not sure i know to prove that there exists a projective transformation that keeps a circle fixed and maps a line to infinity, but i think the inversion model above works fine).
the new triangle will be $A'B'C'$. let $a$ be the tangent to $\omega '$ in $A'$. because $a\cap B'C'$ will be the image of $M$, it follows that $a\| B'C'$, and it follows that $A'B'C'$ is equilateral.
(is it correct ?  :blush: )

another question...not very scientific but more "practical". are the results above "allowed" to be used without proof in a contest ?

I don't see how you prove

[quote="maky"]that there exists a projective transformation that keeps a circle fixed and maps a line to infinity[/quote]

but if you can prove this, this yields a simpler proof of Theorem 1 indeed.

[quote="maky"]another question...not very scientific but more "practical". are the results above "allowed" to be used without proof in a contest ?[/quote]

Ask the organizers. I would expect that it can be used at the IMO, but I am not absolutely sure. Remember that explicitely stating the theorems you use is always better than just mentioning them by (potentially unknown) names.

  Darij

Actually, this lemma that there exists a projective transformation that keeps a circle fixed and maps a given line to infinite one is a problem 30.17 in Prasolov Book (isn't it one of the most important lemmas in projective geometry?). Solution presented there is not very complicated, I've also heard that it is possible to prove this with the inversion, as maky's already mentioned.

i read about the lemma in prasolov`s book as well, and also some friends of mine already knew, and said it was "classical".

my idea with "inversion" (more like a pseudo-inversion) is as follows :
take a circle $\omega$ and a point $T$ outside the plane of $\omega$. now, consider the inversion of pole $T$. it will map $\omega$ to some circle $\omega '$ (it can easily be shown by ptolemy`s theorem).
now denote by $\alpha$ the plane of $\omega$ and by $\beta$ the plane of $\omega '$. a central projection of center $T$ between $\alpha$ and $\beta$ (i`m not sure those are the good words...) will map $\omega$ to $\omega '$ also (because if $A$ is a point of $\omega$, it`s image through the inversion will be $A'$, but $T,A,A'$ are collinear, and $A'\in\beta$, thus $A'$ is it`s image through the projection).
now, if the planes $\alpha$ and $\beta$ are not parallel, there will be a line that will map to infinity (i think it`s called the singular line - and it is the intersection of $\alpha$ with the plane through $T$ parallel with $\beta$).
now, all that`s left is to choose a good $T$. i did not think anymore about it, but knowing the line $d$ and the circle, i don`t think it`s too hard.

also, i think there may be other proofs, because the above are only some ideas of mine.


Let  x1 , x2 , ...,xn  real  numbers . Prove  that
x1/(1+x1^2) + x2/(1+x1^2+x2^2) + ...+xn/(1+x1^2+x2^2+...+xn^2) <sqrtn
Hungarian  tst 2002
Thanks

actually ISL 2001, proposed by Bogdan Enescu, a romanian instructor

Here is what I did:

From Cauchy Schwarz we have:

sqrt(n)sqrt(y1^2+..+yn^2)>=sum(k=1,n) (yk)
Now, we take in the last inequality:
yp=xp/(1+x1^2+x2^2+..+xp^2) for all p =1,n, now we have to prove that:

x1^2/(1+x1^2)^2+...+xn^2/(1+x1^2+..+xn^2)^2<1 and we can see that
for p=1,n we have:
xp^2/(1+x1^2+..+xp^2)^2<=xp^2/((1+x1^2+..+x_(p-1)^2)^2(1+..+xp^2)^2=1/(1+x1^2...+x_(p-1)^2)^2-1/(1+x1^2...+xp^2)^2
and  also we have:
x1^2/(1+x1^2)^2=1-1/(1+x1^2).

now adding all the realition we have what we need!
cheers!


If $ a,b,c$ are the sides of a triangle then prove that:

\[ \frac{b}{c\plus{}1}\plus{}\frac{c}{b\plus{}1} > \frac{a}{a\plus{}1}\]

[quote="moldovan"]If $ a,b,c$ are the sides of a triangle then prove that:
\[ \frac {b}{c \plus{} 1} \plus{} \frac {c}{b \plus{} 1} > \frac {a}{a \plus{} 1}\]
[/quote]

Since:

$ \frac {b}{c \plus{} 1} \plus{} \frac {c}{b \plus{} 1} > \frac {b}{b \plus{} c \plus{} 1} \plus{} \frac {c}{b \plus{} c \plus{} 1} \equal{} \frac {a}{a \plus{} \frac {a}{b \plus{} c}} > \frac {a}{a \plus{} 1}$

we are done !  :P


geometry
rectangle
geometric transformation
rotation
induction
vector
absolute value

From a $n\times (n-1)$ rectangle divided into unit squares, we cut the [i]corner[/i], which consists of the first row and the first column. (that is, the corner has $2n-2$ unit squares). For the following, when we say [i]corner[/i] we reffer to the above definition, along with rotations and symmetry. Consider an infinite lattice of unit squares. We will color the squares with $k$ colors, such that for any corner, the squares in that corner are coloured differently (that means that there are no squares coloured with the same colour). Find out the minimum of $k$.

[i]Proposed by S. Berlov[/i]

Could it be n^2-1?  If that is the answer, I might have found something that looks like a solution and I could post it later, but I don't feel compfortable enough right now.  :oops:

Sorry for my English and my modal-verb-mania.

Mmmm, maybe that was not a very appropriate question in a subforum called "unsolved problems"  :roll:. I'll try not to say too many stupid things.

First of all, given a pair of two different unit squares inside a square of side n other than those two made up from the one in the first row and column and the one in the last row and column or the one in the first column and last row and the one in the last column and first row (from now we will call them [u]korners[/u]  :roll:), there exists a corner which contains both, so they must have different colours. This means that the minimum number of colours needed for a single n-sided square is at least n^2-2, and hence k>=n^2-2.

It is possible to show that for n>2, we have k>n^2-2. Suppose for the sake of contradiction (if there are mistakes in my English, I think it's better that they sound funny) that k=n^2-2; then, by the above observation, two opposite korners of any square of side n in the lattice must be of the same colour. Take such a square and call 1, 2 .. n the colours of the u.squares in the leftmost column and n, n+1 ... n+(n-2), 1 the colours in the rightmost one (reading top-down). If we consider for both columns in the square the 2 whole infinite column in the latttice in which each one is contained, is easy to see that they must be colored in the cycle 1, 2, 3, .... n+(n-2)=2n-2. For instance, the u.s. just below the one colored with n in the left column of the  n-square must have colour n+1, since it is an "opposite korner" of the one coloured with n+1 in the right column. For the same reason, the u.s. just below the one coloured with 1 in the right column must be coloured with 2. It is not difficult to formalize the induction step (:whistling:). Since this argument holds for every pair of columns n-1 squares away from each other, we can state that every column is coloured in a cycle of lenght 2n-2 which repeats (even if "out of phase", can I say?) every n-1 columns (a sort of cycle of cycles).
But if this happened, a colour could not appear in more than two different cycles, because it would mean there exists a u.s coloured with that colour which is at a horitzontal distance strictly less than n-1 from a column whose cycle contains that colour and at a vertical distance less or equal to n-1 (  [((2n-2)+1)/2]  ) from a u.s of the same colour of that column, so being contained in the same corner as another u.s. of the same colour. This gives a total of k=(2n-2)(n-1), but (2n-2)(n-1)>n^2-2 for n>2.

In the case n=2 the minimum is obviously k=2 and can be attained with a chessboard colouring.

For n>2 the minimum k=n^2-1 can be attained in the following way. Our aim is to find a set of n^2-1 u.s. such that the whole lattice can be tesselated (well said?) from it in such a way that for each of the translations transforming one into another of the resulting sets of u.s. happens that either one of its components (vertical or horizontal) exceeds n-1 in absolute value or the absolute value of both their components is exactly n-1. If we colour differently each u.s. of the initial set and colour the other sets in the tesselation in the same way, for every two u.s. of the same colour would happen that either they are hor. or vertic. more than n-1 units away or that they are exactly n-1 units away in each direction; in both cases it would not be possible for them to be in the same corner.
Take a square of side n and  take out its bottom right korner (sorry again). Then apply all the traslations resulting of a linear integer combination of the vectors (n-1, -n+1) and (n, 1): that can be such a  tesselation.

[ I now nothing about the theme, but I suppose the shape fo the initial set is somehow superfluous and that what really matters is the pair of vectors from which combination you divide the lattice into equivelency classes (is it correct in English?). To begin with, its vectorial product: n^2-1. Any advice on where to find more about it? ]

Sorry again for the possible mistakes in language or maths and for such a clumpsy demonstration.

i think you are correct
the answer is indeed $n^{2}-1$. during the contest, the main thing in the problem was to write it down very well explained. i lost two points because i supposed too many things "obvious"

[quote="Mimoide"]
For n>2 the minimum k=n^2-1 can be attained in the following way. Our aim is to find a set of n^2-1 u.s. such that the whole lattice can be tesselated (well said?) from it in such a way that for each of the translations transforming one into another of the resulting sets of u.s. happens that either one of its components (vertical or horizontal) exceeds n-1 in absolute value or the absolute value of both their components is exactly n-1. If we colour differently each u.s. of the initial set and colour the other sets in the tesselation in the same way, for every two u.s. of the same colour would happen that either they are hor. or vertic. more than n-1 units away or that they are exactly n-1 units away in each direction; in both cases it would not be possible for them to be in the same corner.
Take a square of side n and  take out its bottom right korner (sorry again). Then apply all the traslations resulting of a linear integer combination of the vectors (n-1, -n+1) and (n, 1): that can be such a  tesselation. [/quote]

Please someone explain the construction. I don't understand it.

Nobody? I don't think it's clear at all, or even correct...


Could someone provide me a list of the important identities involving binomials? The symmetry identity I know (who doesn't?), but the rest escapes my mind.

Er well the binomial theorem is one..

$ (x\plus{}y)^n\equal{}\sum_{k\equal{}0}^n{n \choose k}x^{n\minus{}k}y^{k}$

Some more (this might be too many of something, meh):

$ \binom{n}{k}\plus{}\binom{n\plus{}1}{k}\equal{}\binom{n\plus{}1}{k\plus{}1}$
$ \binom{n}{0}\plus{}\binom{n}{1}\plus{}\cdots \plus{}\binom{n}{n}\equal{}2^n$
$ \binom{n}{k}\equal{}\binom{n}{n\minus{}k}$
$ \binom{n}{k}$ is the number of ways to choose $ k$ items from a set of $ n$ items.
$ \frac{\binom{2n}{n}}{n\plus{}1}$ (Formula for the Catalan numbers)
See here for more info on Catalan numbers:
[url]http://en.wikipedia.org/wiki/Catalan_number[/url]

Also, another important one is the Hockey Stick Identity:
[url]http://www.artofproblemsolving.com/Wiki/index.php/Hockey_Stick_Identity[/url]

It's more important to learn proof techniques for binomial identities than to learn binomial identities.  For example, a great many binomial identities can be proven just using the generating functions

$ (1 \plus{} x)^n \equal{} \sum_{k\equal{}0}^{n} {n \choose k} x^k$
$ \frac{1}{(1 \minus{} x)^{n\plus{}1}} \equal{} \sum_{k \ge 0} {n\plus{}k \choose n} x^k$

most notably both forms of [url=http://en.wikipedia.org/wiki/Vandermonde%27s_identity]Vandermonde's identity[/url].  (The latter is just the "balls and urns" generating function.)  Differentiating and integrating these identities gets you a lot more identities, but don't forget to make sure you also understand the underlying [url=http://www.mathlinks.ro/weblog_entry.php?t=220351]combinatorial arguments[/url].


calculus
integration
search
calculus computations

1) Show that the integral of sin(x^2)dx from 0 to infinity converges.

It's well known, and has come up plenty of times. Search for the Fresnel integral.


Find an $ x\in\mathbb{Z}$ such that

\[ x^{3}\plus{}4x^{2}\plus{}2x\minus{}2\equiv 0\bmod 625.\]

Start with $ x \equal{} 1$ and apply Hensel's lemma.
I will never understand what the meaning of these ugly problems is...

$ x\equal{}91$

Check mod $ 5$, $ 5^2$ and finally $ 5^3$


find two subspaces $ A$ and $ B$  of $ \mathbb{R}^4(R)$ suchthat $ dim A \equal{} 2$and $ dim B \equal{} 3$and $ dim(A\cap B) \equal{} 1$

[hide="Hint:"] Why is $ A\plus{}B\equal{}\mathbb R^4$ a necessary condition?[/hide]

if $ W_1$ and $ W_2$ are two subspaces of vector space V then $ V$ is direct sum of $ W_1$ and $ W_2$ iff$ W_1\cap W_2\equal{}\{0\}$
i.e. $ V\equal{}W_1\oplus W_2$ if $ W_1\cap W_2\equal{}\{0\}$

What does that have to do with anything?

Try writing down two spaces. Do they work? If chosen randomly, they should work with probability $ 1$.


A ship sails at 20 kilometres/hour in a straight line which is, at its closest, 1 kilometre from a port. A tug-boat with maximum speed 12 kilometres/hour leaves the port and intercepts the ship, leaving the port at the latest possible time for which the interception is still possible. How far does the tug-boat travel?

My solution uses some calculus..
[hide]Let the port be $ P$ and the point of interception $ X$.  Let $ N$ be the point of closest approach of the ship to the port, and $ M$ be the position of the ship when the tug boat leaves $ P$.
Let $ t$ be the time that the tug boat is travelling.

$ PX \equal{} 12t$, $ 1\equal{}12t\cos\theta$
$ MN \plus{} NX \equal{} 20t \Rightarrow MN \equal{} 20t \minus{} 12t\sin\theta$
Substitute in $ t\equal{} \frac{1}{12\cos\theta}$ to get:
$ MN \equal{} \frac{5}{3} \sec\theta \minus{} \tan\theta$

Since the tug leaves at the latest time possible, $ MN$ is minimised, so

$ \frac{d}{d\theta}(MN) \equal{} \frac{5}{3}\sec\theta \tan\theta \minus{} \sec^2\theta \equal{} 0$
$ \sec^2\theta \left (\frac{5}{3}\sin\theta \minus{} 1 \right) \equal{} 0 \Rightarrow \sin\theta \equal{} \frac{3}{5}$ (You can easily check this gives a minimum)
$ PX\equal{}\frac{1}{\cos\theta} \equal{} \frac{1}{\frac{4}{5}} \equal{} \boxed{\frac{5}{4}\text{km}}$
[/hide]

dont think that is nessacary...couldn't u do it w/ sim triangles?


I don't understand what it is. can someone explain it in a clear way?   :)

[url]http://mathworld.wolfram.com/Apothem.html[/url]
As clear as possible, with pretty pictures of happiness.

http://www.mathwords.com/a/apothem.htm

not only a pretty picture, but also how to pronounce it


If $ a,b,c \in (0,1)$ prove that

$ \sqrt {abc} \plus{} \sqrt{(1\minus{}a)(1\minus{}b)(1\minus{}c)} < 1$

because $ a,b,c \in (0,1)$
$ \therefore LHS < \sqrt[3]{abc}\plus{}\sqrt[3]{(1\minus{}a)(1\minus{}b)(1\minus{}c)} \leq 1$  by AM-GM


geometry
geometric transformation
rotation
topology
group theory
advanced fields
advanced fields unsolved

I am asked to find all finite groups $ G$ which can act freely on the circle $ S^1$. I suspect that the answer is "all cyclic groups", but I don't see how to prove this. I thought about using the fact that the action of $ G$ on $ S^1$ must be properly discontinuous, so that $ S^1 \to S^1/G$ is a covering map having $ G$ as the group of covering transformations. But this does not seem to help me to get any further.

Any hints?

you've made it: once you have that $ G$ acts in a properly discontinous way, you have that $ G$ is the group of automorphism of the covering map $ S^1\to S^1/G$, so $ G$ is a quotient of $ \pi_1(S^1) \equal{} \mathbb{Z}$, so it's cyclic. moreover, it can't be infinite, since $ S^1$ is compact and the action can't be properly discontinuous.
on the other hand, rotations by $ 2k\pi/n$ provide a free action of the cyclic group of order $ n$ for every $ n$, and that's the other implication.

Wait, why is it a quotient of $ \pi_1(S^1)$? That's what I want, but I must be missing something there.

hm.. no, actually this works out coverings of $ S^1$, which is the opposite of what we're looking for..
we can reduce ourselves to this case by observing that the quotient is still a compact 1-manifold, so it's $ S^1$ itself.
but i guess there's a more direct way using the fact that $ S^1\setminus {\rm orbit}$ is a union of contractible sets.. i'll try to think about it (maybe).

We have slightly more information than has been pointed out so far. $ S^1\to S^1/G$ is a $ |G|$-fold covering. Does this fact along with $ S^1/G\cong S^1$ give us that the covering map is actually $ z\mapsto z^{|G|}$?

How do I prove rigorously that $ S^1/G$ is homeomorphic to $ S^1$? Which things do I need to prove (in detail) to get this result?

i guess it's classification of manifold of dimension 1 plus the fact that the quotient has to be a manifold.
classification of 1-manifolds should be quite easy, just try it by yourself..


[color=purple] Thua toan the quy vi va cac ban Mathnfriend da chinh thuc haot dong tro lai voi ten mien la [url]http://mathfriend.org[/url] :D  :D
   Cung tham gia nhe ! :D  [/color]

Anh đây!  :lol: Hay đấy , anh vào xem qua rồi! :)

Anh không gặp được chú hôm đó rồi, tiếc quá! Anh sẽ upload CRUX lên mathscope sau! Giọng anh ốm ốm á?  :lol:

[quote="evarist"][color=purple] 19 nay` em ra Ha` Noi va` o het ngay` 20 moi ve` ! Bao gi`o anh ra hoc tiep ? Hi vong khi do anh em minh` co the gap nhau  :D
@ Giong anh nghe qua dien thoai om om rat buon cuoi buon cuoi hon ca giong Nghe An cua thang` ku Quang  :D [/color][/quote]
Hic! Ngày 19 em lại về quê -> Không gặp được anh nói chuyện rồi!


$ a,b,c,d,e$ are real numbers such that

$ a+b+c+d+e=8$
$ a^{2}+b^{2}+c^{2}+d^{2}+e^{2}=16$

What is the largest possible value of $ e$?


advanced fields
advanced fields unsolved

How to find all the integer solutions of $ y^2 \plus{} 21 \equal{} x^3$ use the fact that the ideal class group of $ \mathbb{Q}(\sqrt { \minus{} 21})$ is $ \mathbb{Z}_2\times \mathbb{Z}_2$?

Note that $ \mod 4$ results in $ y$ even, $ x$ odd; $ \mod 21^2$ gives that $ y$ is coprime to $ 21$. Factor as $ (y\plus{}\sqrt{\minus{}21})(y\minus{}\sqrt{\minus{}21})\equal{}x^3$ and see that the ideal $ <y\plus{}\sqrt{\minus{}21} , y\minus{}\sqrt{\minus{}21}> \equal{} <y\plus{}\sqrt{\minus{}21} , 2 \sqrt{\minus{}21}>$ divides $ <y\plus{}\sqrt{\minus{}21} , \sqrt{\minus{}21}> \cdot <y\plus{}\sqrt{\minus{}21} , 2> \equal{} <1>$. Thus the ideals $ <y\plus{}\sqrt{\minus{}21}>$ and $ <y\minus{}\sqrt{\minus{}21}>$ are coprime. This gives that there is an ideal $ \mathfrak a$ such that $ \mathfrak a^3 \equal{} <y\plus{}\sqrt{\minus{}21}>$. But as $ \mathfrak a^3$ is principal and the class group has order coprime to $ 3$, we get that $ \mathfrak a$ is principle, say $ \mathfrak a \equal{} <a\plus{}b\sqrt{\minus{}21}>$.
As all units are $ \pm 1$, this leads us to $ (a\plus{}b\sqrt{\minus{}21})^3 \equal{} \pm (y\plus{}\sqrt{\minus{}21})$.
Comparing coefficients gives $ \pm y\equal{}a^3\minus{}63ab^2, \pm 1\equal{}3a^2b\minus{}21b^3\equal{}3b(a^2\minus{}7b^2)$, the latter being impossible $ \mod 3$, thus there are no solutions.


analytic geometry
topology
advanced fields
advanced fields unsolved

I encounter for generalization of uniformization to calculate fundamental group of this space.
[i]assume that we delete infinite line from $ \mathbb{C}P^2$ so we have$ \mathbb {C}^2$ and we delete complex lines :$ z\equal{}0,z\equal{}1,w\equal{}0,w\equal{}1,z\equal{}w$ in which $ (z,w)$ are coordinate in $ \mathbb{C}^2$ .what is the fundamental group of complement of these lines . [/i]
I know the fundamental group is $ F_2$ semi-direct product $ F_3$ but I don't the action. .

Does anyone know the answer? I think, I have seen it somewhere!


calculus
function
derivative
calculus computations

I am stuck answering this problem and I really need help.

The problem asks for the (f^-1)'(a) for the function f and the real number a.

the function is f(x)=x^3+2x-1 and a=2.

The answer in the back of the book is 1/5.

I can solve other inverse functions but I am stuck with this one. Please help.

$f(g(x)) = x \Rightarrow g'(x)f'(g(x)) = 1 \Rightarrow g'(x) = \frac{1}{f'(g(x))}$

So in your problem we have:

$f^{-1}' (a) = \frac{1}{f'(x)} \text{ where }  f(x) = a $

$x^3+2x-1=2 \Rightarrow x = 1$

$\frac{1}{f'(1)} = \frac{1}{5}$

By the way, you should post this type of question in the calculus computations and tutorials section.

[quote="lianafan"]I do not understand where do you get the 5 from x^3+2x-1.[/quote]

$f(x)  = x^3+2x-1 \Rightarrow f'(x) = 3x^2+2 \Rightarrow f'(1) = 5$

[quote="lianafan"]Can someone explain what is the relationship between the invese and the derivative, please?[/quote]

Read my first post, $f$ and $g$ are inverse functions ($f^{-1} = g$ and $g^{-1} = f$).


number theory
Diophantine equation
number theory proposed

Show there is infinite intengers $ k$ no perfect cubic such that $ a^{3}+b^{3}+c^{3}= k$ admit infinity solutions in $ (a,b,c) \in (\mathbb Z^{*})^{3}$..

[quote="zweig"]Let be $ (a, b, c)\in\mathbb{Z}^{*}$ and $ k$ an intenger $ \neq (-1,0,1)$. Show that it exists some infinite $ k$ such that the equation $ a^{3}+b^{3}+c^{3}= k$ admit infinite solutions in $ \mathbb{Z}$.[/quote]
Try $ (a,b,c,k)=(-x,x,y,y^{3})$, for arbitrary $ x, y \in \mathbb{Z}^{*}$ and $ |y| \ge 2$. :maybe:

And then? I can't understand your remark. :huh: You asked to prove there're infinitely many $ k \in \mathbb{Z}$, $ |k| \ge 2$, such that the Diophantine equation $ a^{3}+b^{3}+c^{3}= k$ has infinitely many solutions in integers $ x, y, z \ne 0$. Is it right? Then the $ 4$-tuple $ (a,b,c,d) = (-x,x,y,y^{3})$, for arbitrary $ x, y \in \mathbb{Z}$ and $ |y| \ge 2$, provides an immediate answer to your question. Still right?

Oh ok, you meant, your solution works ? Hummm  :roll: OMG, you killed my 1st exercice creation  :mad:

And if I say "k not cubic perfect" ? ....  :P

I post my own solution :

[b]Show there is infinite intengers $ k$, no perfect cubic such that $ x^{3}\plus{}y^{3}\plus{}z^{3}\equal{} k$  admit infinity solutions in $ (x,y,z)\in\mathbb{{Z}{}^{*}}^{3}$[/b]

$ (a^{3}\plus{}6n^{3})^{3}\plus{}(a^{3}\minus{}6n^{3})^{3}\plus{}(\minus{}6an^{2})^{3}\equal{} 2a^{9}$ for every $ (a,n)\in\mathbb{{Z}{}^{*}}^{2}$

what ensures the existence of an infinity of $ k$, no cubic perfect, such that the equation admit infinity solutions in $ (x,y,z)\in\mathbb{{Z}{}^{*}}^{3}$


A family consisting of a mother, father, daughter, and three sons participate in a mixed doubles tennis tournament. In mixed doubles, each team has one female and one male. How many different family teams are possible?

This is $ \binom{2}{1} \times \binom{4}{1}\equal{}\boxed{8}$.


trigonometry
inequalities proposed
inequalities

Prove that for every $ x\in \mathbb R$, we have that $ |sinx|\plus{}|sin(x\plus{}1)|\plus{}|sin(x\plus{}2)|>\frac{8}{5}$

[quote="mr.danh"]Prove that for every $ x\in \mathbb R$, we have that $ |sinx| \plus{} |sin(x \plus{} 1)| \plus{} |sin(x \plus{} 2)| > \frac {8}{5}$[/quote]
How about $ x\equal{}0$.  :wink:

I don't understand your idea.
When x=0, sin 0 + sin 1 + sin 2 >1.75>8/5
(sin 1=0.84, sin 2 = 0.9)
maybe you thought about $ \sin 1^o, \sin 2^o$ ?

Could it be improved, i.e. $ 1 \plus{} 2\cos 1\geq LHS \geq 2\sin 1$ (LHS is periodic with the basic period $ \pi$, achieving its minimum in $ \minus{} 1 \plus{} k\pi$, and maximum in $ \minus{} 1 \plus{} \frac {k\pi}{2}$)?


limit
logarithms
real analysis
real analysis unsolved

Determine the values of $x$ such that there exisst a strictly increasing positive sequence of integers $(a_n)_{n\geq 1}$  such that :
     $b_n=\frac{\ln(2^{a_1}+2^{a_2}+...+2^{a_n})}{n}$
converges and $\lim_{n\to\infty} b_n=x$.

Since $a_n\ge n,\forall n\in \mathbb{N}$, $b_n\ge \frac{\ln (1+2+2^2+...+2^{n})}{n}\ge \ln 2$ where $k=a_1$
hence, $x\ge \ln 2$.

We have that $\frac{a_n}{n} \ln 2 \le b_n \le \frac{a_n+1}{n}$, hence by corseting results $\ln 2 \lim \frac{a_n}{n}=\lim {b_n}$. The problem reduces to find those numbers $x$ which can be the limit of a sequence of the form $\ln 2 \frac{a_n}{n}$ with $a_n$ strictly incresing sequence of integers.
For any $x\ge \ln 2$ taking $a_n=[\frac{nx}{\ln 2}]$ we get that $x$ have the desired property.(it is easy to prove that $[\frac{nx}{\ln 2}]\le [\frac{(n+1)x}{\ln 2}]$).

Answer: $x\ge \ln 2$


Hello, this is my first post and i'm really in troubled with one inequality. I think it should not be very complicated for you.
 [code]Verifier correctness of inequality:
2a²+b²+c² >= 2a(b+c)[/code]

I have tried different ways, but well they are all useless, i finish where i started :(

This is a simple application of the Arithmetic-Geometric Mean Inequality, which states that:

$\frac{a + b}{2} \ge \sqrt{ab}$

So applying the inequality to a^2 and b^2, we get:

$\frac{a^2 + b^2}{2} \ge ab$

$\Rightarrow a^2 + b^2 \ge 2ab$

Similarly:

$a^2 + c^2 \ge 2ac$

Therefore, adding up the individual inequalities, we get our desired answer:

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

Note that AM-GM only works for non-negative integers a and b...but a^2, b^2 and c^2 are always non-negative, so the inequality holds.

Try proving the AM-GM inequality.

EDIT:  AH edit!  Alternate solution, cuz I'm a fool:

$a^2 - 2ab + b^2 + a^2 - 2ac + c^2 \ge 0$

$(a-b)^2 + (a-c)^2 \ge 0$

The trivial inequality holds, with equality whenever a = b = c.

EDIT 2:  Yeah thanks.  Caught da error.

Those two solutions are exactly the same. In fact I believe that the proof for AM-GM in 2 variables is simply rearranging the trivial inequality, $x^2\ge 0$.

[quote="lowest_reynolds"]$a^2 - 2ab + b^2 + a^2 - 2ac - c^2 \ge 0$[/quote]
Typo Error. Correct expression is $a^2 - 2ab + b^2 + a^2 - 2ac + c^2 \ge 0$

An another way to prove it :

$LHS - RHS = (a-b)^2 + (a-c)^2$

so $LHS \geq RHS$ with equality iff $a=b=c$

Thank you all, yes now i undestand that it was quite easy :blush:.


I've done complex diagrams in the past using various packages, but for the life of me, I can't figure out how to do a simple diagram of a real line and some open intervals on it  (need it for an elementary handout intended for undergrads).  :(  You know, just a straight line with two arrows pointing towards $\pm\infty$, optionally ticmarks, parentheses that delimit the intervals, and a number under each parenthesis.

I've been searching the web in vain, including information about xfig, gnuplot, and some LaTeX packages.  Obviously, I can draw the whole thing manually in xfig, treat each parenthesis manually as a little arc, etc., but there must be something out there that automates the process...  :?:

Any help is greatly appreciated.

I'd recommend drawing it in a different program and importing it into your $\text{\TeX}$ file. You can definitely do a good job of it with Geometer's Sketchpad if you have it.

I'll check if there's a free (limited) version of Geometer's Sketchpad on the web.

Thanks for your suggestion.

The Eukleides package is the almost perfect solution.  Relatively little Eukleides code can produce much by way of output.  I like the included euktoeps shell script which converts an .euk file to EPS.  Covers all elementary geometric objects, including conic sections, and can compute intersections of line with lines and lines with circles.  Pretty neat.  Relies on PSTricks and some PSTricks-based packages (such as pstricks-add and psteucl).


pigeonhole principle
combinatorics proposed
combinatorics

let $ S \equal{} \{ 1,2,\ldots ,2008 \}$ , find the minimum $ n$ such that every $ n$-element subset of $ S$ ($ S_i$) have two different elements $ a_i$ , $ b_i$ such that all the sums $ a_i \plus{} b_i$ are equal.

We we'll prove that minimum of $ n$ is $ 1005$.
now consider : $ A \equal{} \{1,2,\cdots,n\}$ and $ B \equal{} \{n \plus{} 1,n \plus{} 2,\cdots,2n\}$
now for every two different number in $ A$ we have
$ a \plus{} b\leq2n \minus{} 1$
and for every two different number in $ B$ we have :
$ c \plus{} d\geq2n \plus{} 3$ .
so $ a \plus{} b$ and $ c \plus{} d$ are not equal which is contradiction . Thus we have $ n\geq 1005$ .
now by Pigeonhole theorem we can prove that for every subset with $ 1005$ elements there exist two  different elements $ a$ and $ b$ such that : $ a \plus{} b \equal{} 2009$ and the proof is finished :)


I'm extremely new to this topic
Judd Toone writes:

y = -3 sin (2x+pi/3)
y= 1/x2+1 (1 over xsquared)
y=sinx

Hi my names Judd...
I\'ve been asked to sketch the following graphs of the above problem. I\'ve no idea where to start-Id appreciate any step by step help you can give me -

Well, what do you know about the sine function in particular, or graphing curves in general?


a. proof that the real equation $ x^4 \minus{} x^3 \minus{} 1 \equal{} 0$ has exactly two real
roots.

b. if a and b are the sum and the product of these real roots respectively,
proof that

b < -11/10, a > 6/11.

this is an indian olympiad problem, and the deadline was today. so please, i need the solutions.

i solved the first part, but cannot do the second part.

Same as in the other threads: I want a proof that the deadline is over!


How many positive integers $ n$ satisfy the following condition:
\[ (130n)^{50} > n^{100} > 2^{200}?
\]$ \textbf{(A)}\ 0\qquad \textbf{(B)}\ 7\qquad \textbf{(C)}\ 12\qquad \textbf{(D)}\ 65\qquad \textbf{(E)}\ 125$

[hide="Solution"]Take the 50th root of everything:
$ 130n>n^2>16$
Looking at the first part: $ 130>n$.
Now at the second: $ n>4$.
So we have $ n\in[5,129]$, and the answer is $ 129\minus{}5\plus{}1\equal{}\boxed{\textbf{(E)}\ 125}.$[/hide]


This message just got sent to ARML coaches:

[quote="Steve Adrian"]To all ARML Coaches:

The SAT's will be given on June 7, 2008  and ARML is May 30 and 31, so students will not need to have an Alternate Test Day.
[/quote]

Wait... I thought ARML was supposed to be on the first Saturday in June, which, for 2008, would be June 7. The date was even marked in red on the calendar I got at ARML this year.

Hey- there's a chance I might be able to escape Washington and be there in Las Vegas. One week earlier on the academic calendar makes a big difference.

Replying to sq53e, a second email message:

[quote]To all ARML Coaches:

Please accept our apology for the error on the D.E Shaw Calendar that was handed out at the cmpetition. ARML 2008 will be on May 30, 31, 2008. Please put this date into your plans. Thanks.

Steven L. Adrian
ARML Executive Director
[/quote]

[quote="Sly Si"]It's not actually the first Saturday in June.  I think it's the first weekend after Memorial Day.  It was on the same dates in 2003, if I remember right.

This also means that my chances of going go from "uncertain" to "certain", as it no longer falls during my final exam week.[/quote]

How do you plan on getting there? Now that I'll be considerably closer, I think it'd be fun to stop by :).

[quote="xxazurewrathxx"]It's about time they stop conflicting with the SATs.[/quote]

Was that ever a real problem? I just asked my teacher to tell them i'm going to arml, and then I just did the sat on a different day (and it wans't as early in the morning, so it was ultra chill)

[quote="Sly Si"]You can go with SFBA, as I've been doing.  SUMO (the Stanford math club) provides coaches to help out with practices, and anyone who wants to can go as a coach to Vegas for the tournament.  Usually, students don't go because ARML falls the weekend before finals start, but I'd love to see that change.[/quote]

yooooooooo

is this hard to qualify for or whatever? I'd so totally be up for it, but I haven't had any relations with the SFBA group or whatever and I dunno how qualified I'd be to help. Is it just anyone in SUMO (Actually, I think I'm leaving for stanford whenever the school FINALLY starts so I should probably look into this) that wants to do this can do it?

Anyone from SUMO can, as far as I know.  In the past, I've been the only one, but that can change.  We might have to be careful about space on the bus, but Ted's policy has long been that anyone who wants to go to ARML (including SUMO people) can go.

Oh man, that's cool, I'm so totally up for it next ARML. Plus, I get to chill with the Alabama team at UNLV, which is another benefit. And the lack of conflict with final exams is nice (although I bet I'd still be up for it even if it did conflict). What are you required to do if you decide to go?

May 30-31? Sweet. That ups my chances of going from 20% to 95%.

[quote="Elemennop"][quote="xxazurewrathxx"]It's about time they stop conflicting with the SATs.[/quote]

Was that ever a real problem? I just asked my teacher to tell them i'm going to arml, and then I just did the sat on a different day (and it wans't as early in the morning, so it was ultra chill)
[/quote]

Yep, they don't let us take alternate SATs in Florida. That is, none of the schools will proctor it.  :(

We lose around 5-6 of our top students every year to the SAT  :dry: , the rest of us have to plan our SATs around it.

people would actually care more about taking an sat then going on a fun weekend trip

wtf

it's like they dont WANT to have fun

I know, right? Well, it's pretty common among the Floridian math kids to take the July SAT because everyone wants to take advantage of the fact that they just finished studying for finals/APs/IBs to finish off their SAT II as well.

It's common everywhere- practically every team loses people to the SAT every year. There just aren't that many weekends to go around, with the requirements of being late enough that the college hosts have room and early enough that the high school kids haven't scattered for summer.

I only went to ARML myself once, at the end of my 12th grade year. The year before, I took the SAT II instead, and I didn't even know about ARML in 9th and 10th grade.

Ah well, our problem is that we exclusively lose our very top people, and we don't have many top students to go around in Florida, unfortunately.  :(

I'm gonna be trying out for it this year, and cause I'm only in 10th grade I won't have to deal with SATs (I hope.) Really annoying that they would have it conflict though, and glad that it doesn't this year.


A target consists of four concentric
circles with diameters 6 inches,
8 inches, 10 inches and 12 inches
as shown. Given that a dart is
thrown at random and hits the
target, what is the probability that
it hits a shaded region? Express
your answer as a common fraction.

[asy]size(100); import graph; fill(Circle((0,0),6),gray(.6));
clip(Circle((0,0),3) ^^ Circle((0,0),4) ^^ Circle((0,0),5),fillrule(1));
draw(Circle((0,0),3) ^^ Circle((0,0),4) ^^ Circle((0,0),5) ^^ Circle((0,0),6));[/asy]

The area of the shaded region is $ 5^2\pi\minus{}4^2\pi\plus{}3^2\pi\equal{}18\pi$ and the area of the whole board is $ 6^2\pi\equal{}36\pi$, so the probability is \[ \frac{18\pi}{36\pi}\equal{}\boxed{\frac{1}{2}}\]


Suppose the true weight of a standard weight is 10 grams. It is weighed twice independently. Suppose that the first measurement is a normal random variable X with E(X)=10g and SD(X)=0.2g, and that the second measurement is a normal random variable Y with E(Y)=10g and SD(Y)=0.2g.

a) Compute the probability that the second measurement is closer to 10g than the first measurement.
b) Compute the probability that the second measurement is smaller than the first, but not by more than 0.2g.

Any help is appreciated! Thanks.

My suggestions:

(a) There are three situations:  Either (1) $ X$ is closer than $ Y$, (2) $ Y$ is closer than $ X$, or (3) they're both the same distance away.  Which is more likely, (1) or (2)?

(b) Let $ Z\equal{}X\minus{}Y$.  What is the distribution of $ Z$?  How can you describe the event you're looking for in terms of $ Z$?


Fifteen balls are numbered $ 1\minus{}15$. Four balls are then selected at random without replacement. What is the probability that the sum of the numbers on those four balls is odd? Express your answer as a common fraction.

We must have either 3 odd balls and one even or 1 odd and three even.  Calculate the probabilities separately, add them, and we get $ \boxed{\frac{32}{65}}$.


algebra polynomial trigonometry Vieta calculus ratio

[hide="Is it this simple? :huh: ...."]$a+b+c=0$

$ab+bc+ca=-3$

$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$

$\implies a^2+b^2+c^2=6$  which is what we have to prove(just add up the three things we have to prove)[/hide]

Something must be wrong....

[quote="kunny"]Let $a,b,c$ be the roots of the cubic equation  $x^3-3x+1=0.$ If $a<b<c,$ then prove that we have $a^2=c+2,\ b^2=a+2,\ c^2=b+2.$[/quote]

The polynomial with roots that are 1 less than the roots of $x^3-3x+1=0$ is $x^3+3x^2-1=0$. Thus $(a-1)(b-1)(c-1)=1$. Multiplying both sides by $(a+1)$, we have $(a+1)(a-1)(b-1)(c-1)=a+1$. But $(a+1)(a-1)=a^2-1=c+1$, so we have $(c+1)(c-1)(b-1)=a+1$. Since $c^2-1=b+1$, we have $(b+1)(b-1)=a+1$, or $b^2=a+2$, oh shoot this is circular.

[quote="chess64"][quote="kunny"]Let $a,b,c$ be the roots of the cubic equation  $x^3-3x+1=0.$ If $a<b<c,$ then prove that we have $a^2=c+2,\ b^2=a+2,\ c^2=b+2.$[/quote]

The polynomial with roots that are 1 less than the roots of $x^3-3x+1=0$ is $x^3+3x^2-1=0$. Thus $(a-1)(b-1)(c-1)=1$. Multiplying both sides by $(a+1)$, we have $(a+1)(a-1)(b-1)(c-1)=a+1$. But $(a+1)(a-1)=a^2-1=c+1$, so we have $(c+1)(c-1)(b-1)=a+1$. Since $c^2-1=b+1$[/quote]

You used the result. :(

[quote="kunny"]Let $a,b,c$ be the roots of the cubic equation  $x^3-3x+1=0.$ If $a<b<c,$ then prove that we have $a^2=c+2,\ b^2=a+2,\ c^2=b+2.$[/quote]

let $\sqrt{y}=x$, then

$\sqrt{y}(y-3)=-1$
$y^3-6y+9y-1=0$

(the roots of this are $a^2$, $b^2$, $c^2$)

let $z-2=x$

$z^3-6z^2+12z-8-3z+6+1=0$
$z^3-6z^2+9z-1=0$

(the roots of this are $a+2$, $b+2$, $c+2$)

now the coefficients of the polynomials are the same, therefore they have the same roots, clearly, we don't have that $a^2=a+2$, but since they have the same roots, they have to pair up disjointly,

testing some values, and using the intermediate value theorem, $-2<a<-1$, $0<b<1$, and $1<c<2$

looking at these gives the result

Also:
[hide]One can see that $a= 2\cos (160^o),b= 2 \cos (80^o) , c = 2 \cos (40^o)$ :) [/hide]

wow...i was actually thinking trig substitution, but i did not explore that solution that much...but i don't see how you would know that without seeing the problem before

If you are looking for a computational solution, use the Tschirnhaus and then the Vieta substitution.

[quote="4everwise"][quote="Gyan"][quote]but i don't see how you would know that without seeing the problem before [/quote]

Just Let   $x=a \cos \theta$
we see  $a^3 \cos ^3 \theta - 3 a \cos \theta +1 = 0$
Now hope for  $\frac{a^3} {3a} = (4/3)$ so get  $a=2$  and you get nice formula for $cos (3\theta) = -(1/2)$ rest  is easy. Works [i]every [/i]time for any cubic equation  as long as you know $\cos (3\theta) = 4 \cos ^3 \theta -3\cos \theta$  ...and all the roots are real  :)[/quote]Nice. :) Now I know trig substitution for algebra and calculus! :P

Is there a better way to find the value of a, or do you just hope that it's some nice value?[/quote]

"hope for" was said as  in toungh in cheek. To find 'a' is as simple as anyting.   ($a^2/3$  is known  - just the ratio of coefficents of  $X^3$ and $X$ so it's a piece of cake.
(Other thing is to know, how to convert any cubic equation (by substitute y=x+k and thus finding the value of k) so that $x^2$ term becomes zero - but that  method is given in any nice book)

[quote="kunny"]Let $a,b,c$ be the roots of the cubic equation  $x^3-3x+1=0.$ If $a<b<c,$ then prove that we have $a^2=c+2,\ b^2=a+2,\ c^2=b+2.$[/quote]

Consider backward, since c is the root of the given cubic equation, we have $c^3-3c+1=0,$ thus .....


How many ways can $ \$13$ be made using only quarters and/or
dimes?

Each 50 cents can be made using either 2 quarters or 5 dimes. in $ \$13$ there are 26 50-cent portions.  Dimes can make up any number from 0-26 of these portions.  There are 27 integers from 0-26, so the answer is 27.


A cyclist rides his bicycle over a route which is $\frac{1}{3}$ uphill, $\frac{1}{3}$ level, and $\frac{1}{3}$ downhill. If he covers the uphill part of the route at the rate of $16$ miles per hour and the level part at the rate of $24$ miles per hour, what rate in miles per hour would he have to travel the downhill part of the route in order to average $24$ miler per hour for the entire route?

[hide]
For this we need the harmonic mean of $16,24,x$ to be 24.
So $\frac{\frac{1}{16}+\frac{1}{24}+\frac{1}{x}}{3}=\frac{1}{24}$
Multiplying by 3 gives:
$\frac{1}{16}+\frac{1}{24}+\frac{1}{x}=\frac 18$
Simplifying gives:
$\frac 1x=\frac{1}{48}$
So $x=48$[/hide]


Prove that there are infinitely many positive integers $n$ such that $n^2|3^n-1$.
And find the values of the first five $n's$. :)

[quote="Megus"]I think it's not correct. To have desired divisibility, we must have:

$r \phi (n^2) = n$ for some natural $r$, suppose $n=\prod p_i ^{\alpha}$, then:

$r \prod (p_i-1)p_i ^{2 \alpha -1} = \prod p_i ^{\alpha}$
$r \prod (p_i-1)p_i ^{\alpha -1} = 1$ which is clearly impossible to obtain (only for $n=2$, $r=1$)[/quote]
I don't agree with you. :D Let $n$=1,2,4,20... clearly my problem holds. :D

If $n$ is such a number, then so is $\frac {3^n-1}n$.

[quote="Megus"]I think it's not correct. To have desired divisibility, we must have:

$r \phi (n^2) = n$ for some natural $r$, suppose $n=\prod p_i ^{\alpha}$, then:

$r \prod (p_i-1)p_i ^{2 \alpha -1} = \prod p_i ^{\alpha}$
$r \prod (p_i-1)p_i ^{\alpha -1} = 1$ which is clearly impossible to obtain (only for $n=2$, $r=1$)[/quote]

It is correct that $n^2|3^{\phi(n^2)} - 1$ but that doesn't mean that $\phi(n^2)$ is the smallest number $x$ such that $3^x-1$ is divisiable with $n^2$

example: $100|9^{\phi(100)}-1=9^{40}-1$, but also $100|9^{10}-1$

So:  $\phi(n^2)|n$ -> $n^2|3^n-1$ BUT
from this  $n^2|3^n-1$ it doesn't implies $\phi(n^2)|n$

   hope you get it

M@re


calculus
integration
trigonometry
calculus computations

Evaluate $ \int_{\frac{\pi}{4}}^{\frac {\pi}{2}} \frac {1}{(\sin x \plus{} \cos x \plus{} 2\sqrt {\sin x\cos x})\sqrt {\sin x\cos x}}dx$.

its tough. even the standard tricks of letting the integral be I and substituting $ u \equal{} \frac {\pi}{2} \minus{} x$ do not seem to work. one can show that the given integral is equal to

$ \int^{\frac {\pi}{4}}_{0} \frac {1}{(\sin x\plus{}\cos x \plus{} 2\sqrt {\sin x\cos x})\sqrt {\sin x\cos x}}$

[quote="kunny"]Evaluate $ \int_{\frac {\pi}{4}}^{\frac {\pi}{2}} \frac {1}{(\sin x \plus{} \cos x \plus{} 2\sqrt {\sin x\cos x})\sqrt {\sin x\cos x}}dx$.[/quote]

$ \int^{\frac {\pi}{2}}_{\frac {\pi}{4}} \frac {1}{(\sin x \plus{} \cos x \plus{} 2\sqrt {\sin x\cos x})\sqrt {\sin x\cos x}} dx \\
\equal{} \int^{\frac {\pi}{2}}_{\frac {\pi}{4}} \frac {1}{\sin^2 x(1 \plus{} \cot x \plus{} 2\sqrt {\cot x})\sqrt {\cot x}} dx \\
\equal{} \int^{\frac {\pi}{2}}_{\frac {\pi}{4}} \frac {\csc^2 x}{ (1 \plus{} \cot x \plus{} 2\sqrt {\cot x})\sqrt {\cot x}} dx \\
\equal{} \int^{0}_{1} \frac { \minus{} dz}{ (1 \plus{} z \plus{} 2\sqrt {z})\sqrt {z}} \\
\equal{} \int^{1}_{0} \frac {2du}{ (1 \plus{} u^2 \plus{} 2u) }$

$ \equal{} 1$

  :)   :)

hello, do you mean that $ {\sin^2(x)(1+\cot(x)+2\sqrt{\cot(x)})\sqrt{\cot(x)}=(\sin(x)+\cos(x)+2\sqrt{\sin(x)\cos(x)}\sqrt{\sin(x)\cos(x)}}$? If so, then you can substitute $ t=\cot(x)$.
Sonnhard.


Let $ f_m(x)\equal{}a(m^2\plus{}\frac{1}{m^2})x^2\plus{}b(m^2\plus{}\frac{1}{m^{\frac{3}{2}}})x\plus{}2c\equal{}0$
($ m>0$)
Prove if $ f_1(x)\equal{}0$ have solution then $ f_m(x)\equal{}0$ have solution

It is not true.
For example $ f(x)\equal{}2x^2\plus{}3x\plus{}1$
$ f_(1)\equal{}0\Longleftrightarrow x\equal{}\{\minus{}1,\frac{\minus{}1}{2}\}$
But the $ f(x)\equal{}\minus{}1,f(x)\equal{}\frac{\minus{}1}{2}$ has no solution.

[quote="TTsphn"]It is not true.
For example $ f(x) \equal{} 2x^2 \plus{} 3x \plus{} 1$
$ f_(1) \equal{} 0\Longleftrightarrow x \equal{} \{ \minus{} 1,\frac { \minus{} 1}{2}\}$
But the $ f(x) \equal{} \minus{} 1,f(x) \equal{} \frac { \minus{} 1}{2}$ has no solution.[/quote]
in my mind the wrong is what you wrote. It comes me like you aren't informed in sufficiency.

[quote="mufc"]Let $ f_m(x) \equal{} a(m^2 \plus{} \frac {1}{m^2})x^2 \plus{} b(m^2 \plus{} \frac {1}{m^{\frac {3}{2}}})x \plus{} 2c \equal{} 0$
($ m > 0$)
Prove if $ f_1(x) \equal{} 0$ have solution then $ f_m(x) \equal{} 0$ have solution[/quote]

We have $ \delta_m\equal{}b^2(m^2 \plus{} \frac {1}{m^{\frac {3}{2}}})^2\minus{}8ac(m^2 \plus{} \frac {1}{m^2})$
If $ ac\leq 0$, all equations $ f_m(x)\equal{}0$ have real roots.
If $ ac>0$, we have $ delta_1\equal{}4(b^2\minus{}4ac)$ and so $ b^2\geq 4ac>0$ and it's enough to show that $ (m^2 \plus{} \frac {1}{m^{\frac {3}{2}}})^2\geq 2(m^2 \plus{} \frac {1}{m^2})$

And this last inequality is always true for $ m\geq 2$ since it is $ m^4\plus{}2\sqrt{m}\plus{}\frac{1}{m^3}\geq 2m^2 \plus{}\frac{2}{m^2}$ and we have $ m^4>2m^2$ and $ \sqrt{m}>\frac{1}{m^2}$

And hence the result : $ \delta_1\geq 0$ implies $ \delta_m\geq 0$

I think it's true for all $ m\geq1$ because
we have $ m^4\plus{}1\geq 2m^2$ so we only to prove that :$ 2\sqrt{m}\plus{}\frac{1}{m^3} \minus{}1\geq\frac{2}{m^2}$
                                                                                $ \Leftrightarrow (t\minus{}1)(2t^6\plus{}t^5\plus{}t^4\plus{}t^3\plus{}t^2\minus{}t\minus{}1)\geq 0 (t\equal{}\sqrt{m}\geq1 )$


AMC USA(J)MO USAMO function inequalities logarithms calculus

1) Do you have to show if a function is increasing or decreasing? Say, for the inequality problem on #74, they ask you to prove $a^ab^bc^c\geq(abc)^{\frac{a+b+c}{3}}$, at one point you have to use AMGM within a natural logarithm to get $\ln (\frac{a+b+c}{3})\geq\ln (abc)^{\frac{1}{3}}$; do you have to show that the function $\ln x$ is a an increasing function (by calculus or otherwise), or can you simply state it is increasing and thus the step is valid?

2) Homogenizing...I've been wondering so much about when you can use this. Basically, can someone just give me the low down on when, how, and to what extent you can homogenize an inequality.

[quote="Elemennop"]1) Do you have to show if a function is increasing or decreasing? Say, for the inequality problem on #74, they ask you to prove $a^ab^bc^c\geq(abc)^{\frac{a+b+c}{3}}$, at one point you have to use AMGM within a natural logarithm to get $\ln (\frac{a+b+c}{3})\geq\ln (abc)^{\frac{1}{3}}$; do you have to show that the function $\ln x$ is a an increasing function (by calculus or otherwise), or can you simply state it is increasing and thus the step is valid?
[/quote]

1. If it's an elementary function, like ln x or x^2, to my knowledge you're allowed to simply state it's increasing.

2. If it is a more complicated function, and you can show it's increasing without using calculus, show it's increasing.

3. If you can't show it's increasing without taking derivatives, it's  up to you: you might lose points if you state that the function is increasing without proof, and you might lose points if you use calc and mess up (the graders don't take off points if you use calculus correctly, but they don't like it either).

[quote="Elemennop"]2) Homogenizing...I've been wondering so much about when you can use this. Basically, can someone just give me the low down on when, how, and to what extent you can homogenize an inequality.[/quote]

When all the variables in the inequality are completely interchangeable (or one might call it "indistinguishable"), you can do this. For example, if you had something like

\[ wxyz+\ln(w)+\ln(x)+\ln(y)+\ln(z)+e^{wxyz}-w-x-y-z \\ \ge \tan{(\sin(w)+\sin(x)+\sin(y)+\sin(z)-w^w-x^x-y^y-z^z)} \]

, all the variables are indistinguishable/interchangeable. So, seeing as those four real numbers must have some order, you can just say $w\gex\gey\gez$, and it follows that all cases that don't follow this order can just be found by interchanging variables.

Say WLOG when you can, save some fractions of trees.

Hm? That's not homogenizing... basically if you have a function $f(x_1,\dots , x_n)$ and we replace $x_1,\dots , x_n$ with $kx_1, \dots kx_n$ for some constant (so we're just scaling) then if the inequality is equivalent (i.e. so originally, all the terms are the same degree if $f$ is some polynomial), we can say stuff like $x_1 + \cdots + x_n = 1$ etc

For instance, consider $a,b,c > 0$, $\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} \geq \frac 32$... everything's the same degree so we can WLOG $a+b+c = 1$ for instance.  But we can't do the same thing on the inequality $a^3+bc+b^2c^2 \geq 3abc$ (why?)

Can someone give an example of when homogenizing is useful or necessary?

It's never "necessary," but for example in USAMO 2003/5 it could make stuff easier to work with.

[quote="matt276eagles"]So, how do you homogenize in a proof?  Can you just write, "Without loss of generality, $a+b+c=1$"?  Or do you have to say, "By homogeny..."  Do you actually have to show that if you replace $a$ with $ka$ and so on, then the inequality is the same?[/quote]

I think you can just say "Since the inequality is homogenous, WLOG let ..."

[quote="pieterminate"]Question: can we use formulas or facts about things (e.g.  centroid of triangle, cyclic polygons, etc.....)?    :roll:[/quote]

Well as long as it's well-known(meaning it's on mathworld)...

[quote="beta"]
[quote="pieterminate"]Question: can we use formulas or facts about things (e.g.  centroid of triangle, cyclic polygons, etc.....)?    :roll:[/quote]

Well as long as it's well-known(meaning it's on mathworld)...[/quote]

And when doing proofs, can we "skip" a few steps?
As in can we go directly from $(x+y)^3$ to $x^3+3xy^2+3x^2y+y^3$?
Or factor $x^2+21x+110$ without using the q.f.?

No. You must apply one and only one axiom at each step.  :D

Seriously... the USAMO graders are math professors, not your elementary school teacher (no offense to those elementary school teachers who are very well qualified ;)). Anything you can do in your head without thinking, they probably can too.

Well, ok.
I thought that we were to assume that they were a normal person or something.
 :)

you can't skip steps in your reasoning, because then your proof becomes incomplete.

but algebra and simplification can be shortened (and actually should be shortened to prevent giving your graders a headache) a bit.

[quote="Elemennop"]Alright so I know when to homogenize; if it's able to be homogenized, can you do ANY symmetric sum? Say, $f(ka,kb,kc)=k^2f(a,b,c)$; could we say $a+b+c = 2$, $ab+ac+bc=3$, or $abc =1$?[/quote]

From what I understand, it has to be $f(ka,kb,kc)=f(a,b,c)$. See how the k's cancel out in THazn1's earlier example.

[quote="Elemennop"]Alright, yea, I understand the condition, but can you do any symmetric sum?[/quote]

Yes, I believe you can. Because if you have $ab+bc+ca = k$ for example, simply divide each of $a,b,c$ by a factor of $\sqrt{\frac{k}{3}}$, and you will have $a'b'+b'c'+c'a' = \frac{3}{k}(ab+bc+ca) = 3$. But it's homogeneous so if you prove the inequality for $a', b', c'$ you will have proved it for $a,b,c$ as well.

EDIT: By the way, calc rulz or a mod, do you think you can get rid of that huge LaTeX string that's extending the window in Post #3? It's sort of annoying. Thanks!

[quote="Elemennop"]Alright so I know when to homogenize; if it's able to be homogenized, can you do ANY symmetric sum? Say, $f(ka,kb,kc)=k^2f(a,b,c)$; could we say $a+b+c = 2$, $ab+ac+bc=3$, or $abc =1$?[/quote]

Only if the problem asks for $f(a, b, c) \ge 0$ or $f(a, b, c) \le 0$.  For example, if $f(a, b, c) = a^2 + b^2 + c^2 - ab - bc - ca$ and you must prove $f(a, b, c) \le 0$.

Umm... homogeneous means all terms (on both sides) have the same degree (0 can be any degree). So that means you can do whatever you want, like $a+b+c=1, abc = 1, a^2 + b^2 + c^2 = 1$, etc. Heck, you should even be able to go $a^2 + b^2 + c^2 + 2abc = 1$ in most cases.

No, you can't use homogeneity on every symmetric sum. The sum also has to be HOMOGENEOUS.

Homogeneous: all degrees of the variables are the same
Symmetric: you can permute the variables.

For example, in

$a^3+b^3+c^3-ab-bc-ca \ge 0$

you CAN NOT wlog that $a+b+c = 1$ even though it's symmetric.

No, I think he was talking about letting WLOG $ab+bc+ca=1$ instead of $a+b+c=1$ when the inequality is homogenous.

I think that might be ok as long as every term has a factor of $abc$ in it? I don't think it's valid in every case.


If given a cube, find the angle, theta, formed by the diagonal of the cube and the diagonal of a face of the cube.  Also, find the angles, alpha 1 and alpha 2, formed by the the diagonals of a cube.  Use "x" as the length of a side of the cube.

[hide="Solution"]1. The angle is simply one angle of a $ x \minus{} x\sqrt{2} \minus{} x\sqrt{3}$ right triangle. The angle is therefore:
\[ \sin^{\minus{}1} \frac{x}{x\sqrt{3}}\equal{}\boxed{\sin^{\minus{}1} \frac{\sqrt{3}}{3}} \approx \boxed{35.26^\circ}\]2. Drawing a diagram, we can see that:
\[ \alpha_1\equal{}2 \cdot \sin^{\minus{}1} \frac{\frac{x}{2}}{\frac{\sqrt{3}}{2}}\equal{}\boxed{2 \sin^{\minus{}1} \frac{\sqrt{3}}{3}} \approx \boxed{70.53^\circ}\]Since $ \alpha_1$ and $ \alpha_2$ are supplementary:
\[ \alpha_2\equal{}\boxed{\pi\minus{}2 \sin^{\minus{}1} \frac{\sqrt{3}}{3}} \approx \boxed{109.47^\circ}\][/hide]

[quote="123456789"][hide="Solution"]1. The angle is simply one angle of a $ x \minus{} x\sqrt {2} \minus{} x\sqrt {3}$ right triangle. The angle is therefore:
\[ \sin^{ \minus{} 1} \frac {x}{x\sqrt {3}} \equal{} \boxed{\sin^{ \minus{} 1} \frac {\sqrt {3}}{3}} \approx \boxed{35.26^\circ}
\]
2. Drawing a diagram, we can see that:
\[ \alpha_1 \equal{} 2 \cdot \sin^{ \minus{} 1} \frac {\frac {x}{2}}{\frac {\sqrt {3}}{2}} \equal{} \boxed{2 \sin^{ \minus{} 1} \frac {\sqrt {3}}{3}} \approx \boxed{70.53^\circ}
\]
Since $ \alpha_1$ and $ \alpha_2$ are supplementary:
\[ \alpha_2 \equal{} \boxed{\pi \minus{} 2 \sin^{ \minus{} 1} \frac {\sqrt {3}}{3}} \approx \boxed{109.47^\circ}
\]
[/hide][/quote]

Thanks!! It helped lots! I was stuck on it for two weeks.


calculus
integration
function
logarithms
symmetry
algebra
partial fractions

I have this function which i need to find the integral of. However, i believe i will have to use both integration by parts and substitution. Can you tell me / show me how to start.

$\int_0^{\sqrt{\pi}}x\cdot \cos(x^2)$

Thx in advance...

It's  $[(f \circ g)(x)]' =...$  so you just need to multiply by $2$  .  :)

Oh my, why did i ask about that. That was clearly obvious. I, for one reason or another, forgot everything about the fact that i needed a $\cdot$ as i would need to find $(x^2)'$. Ok, my bad, way too obvious now. Thx though.

However - maybe you can help me prove that this is correct? I have never had any big succes in proving stuff. I see it works if i plot in numbers, but that isn't enough to prove it.

$x=e^{\ln10\cdot \log x}$

And after that:

$\ln x=\ln10\cdot \log x$

You just need to use the change of base for logarithms:

$\log_a(b) = \frac{\ln(b)}{\ln(a)}$

When you have such a rule in your head it is no problem, but i can't recall i have ever heard it before - maybe i have, or at least i have now :) Thx Xevarion.

Just one more question for today, and i should have enough knowledge to go on with some tasks for the rest of the day.

$\int_2^3\frac{x}{1-x^2}dx$

How can i solve this? Well, i would first of take out $x$ from the nominator. Then i have:

$\int_2^3x\cdot (1-x^2)^{-1}dx$

Unfortunately, it is not possible to find the original function when you have something in the power $-1$.

But it is again $[(f \circ g)(x)]' =...$
You must multiply by $-2$ now :)  .

But you can also do the following :

$\frac{x}{1-x^2} = - \frac{1}{2} \cdot \left( \frac{1}{1+x} + \frac{1}{x-1}\right)$  :)

The latter can only be used on the simpler fractions so i will try and solve this with the first method you mentioned. Thx for the simpler method though, but i prefer methods that can be used to all fraction integrations ;) But how do you do that? If you want to do a $[(f \circ g)(x)]'$ on this one, i believe you will end up with a power of $-1$. I have not yet been shown a way of handling fractions - the textbook has only stated integration by parts and substitution, but i believe neither of the two can help me entirely though this problem.

$\int \frac{1}{x} dx = \ln(x) + C$

Do you see what composition Ramanujan was suggesting now?  :)

You can do partial fractions and then you have two simpler logs, if you prefer that method.

There's no general approach for integrating rational functions. These are both good ones to keep in mind.

edit: btw, Ramanujan, I like your signature. Is that the Dalai Lama?

Ok it is $(f \circ g)(x) = \log(x^2-1)$ but you cn solve it by substitution. Set $w = x^2-1$ .  :)

[Edit : I didnt saw Xevarion post ]

[quote="Xevarion"]
 btw, Ramanujan, I like your signature. Is that the Dalai Lama?[/quote]

Yes it is  :)

[quote="Sporally"]:( I think i will need to ask for a walkthrough on this one. I get the point partly, but unfortunately, partly isn't enough here. If you can give me a walkthrough on this one i believe i will be able to relate it to later integration problems containing fractions.[/quote]

What is your problem... ???  :?

$w = x^2-1$  .Then $2x \, dx = dw$ so the integral is $- \frac{1}{2} \int_3^8\frac{1}{w}dw = - \frac{1}{2}[ \ln(w)]_{3}^8$ .

 :)

It is $1-x^2$, not $x^2-1$, but that only changes the bounds. But except that, where did the $x$, which was the nominator, go? That was the problem. I got myself $\int x\cdot (1-x^2)^{-1}$, but then i would have both integration by parts and substitution. Or is there an explanation why this $x$ can be removed?

OK, that was my bad once again. I see where i was wrong, and i have solved the problem with $x$ now. However, the bounds are negative, so i can't put in the bounds since $w>0$, but it is not, so i can't do $\ln(w)$. So that's a problem, right?

$\ln(|w|)$

Think about the symmetry of the $\frac{1}{x}$ function. :-)

[quote="Sporally"]It is $1-x^2$, not $x^2-1$, but that only changes the bounds. [/quote]

This is a property of integral : $\int_{a}^{b} c \cdot f(x) \,dx = c \cdot \int_{a}^{b} f(x) \,dx$ for constant $c$ .

So

$\\ \int_2^3\frac{x}{1-x^2} \, dx = \int_2^3 (-1) \cdot \frac{x}{x^2 -1} \, dx = (-1) \cdot \int_2^3 \frac{x}{x^2 -1} \, dx = \\ \\ \\ = -\frac{1}{2} \cdot \int_2^3 \frac{2 x}{x^2 -1}\, dx = - \frac{1}{2} \cdot \int_2^3 ( \ln (x^2 -1 ))' \, dx = - \frac{1}{2} [ \ln (x^2 -1 )]_{2}^{3}=...$

and there is not any problem now...  :)

Yeah, ignore what I said, listen to Ramanujan.


group theory
abstract algebra
LaTeX
superior algebra
superior algebra solved

Let N be a normal subgroup of index m in G. SHow that a^m in N for all a in G

This problem is very simple. Note that for each $a\in G$ we have: \[(aN)^{|\frac{G}{N}|}= N \Rightarrow (aN)^{m}= N \Rightarrow a^{m}N = N \Rightarrow a^{m}\in N\]

again I would like to point out if I may  :)  that latex is not that hard, just put a^m between dollar signs

: $a^{m}$

[quote="-oo-"]This claim holds for every subgroup $N$. What about a proof? :)[/quote]

NO, claim do not holds for every subgroup $N$. Suppose $H=\{(),(1,2)\}\leqslant S_{3}.$ $H \ntrianglelefteq S_{3}$ and $(1,3)^{3}\notin H.$


iam a new membre and i ask u to give me a site of chinise olympiadse"
100000000000000000000000 thanks


A negative number exceeds its reciprocal by 1.5. What is the number, expressed as a common fraction?

Let this number be $ x$. We can then say \[ x \minus{} \frac{1}{x} \equal{} 1.5\] If we multiply both sides by $ x$ we get \[ x^2 \minus{} 1 \equal{} \frac{3x}{2}\] Multiplying by 2 then gives \[ 2x^2 \minus{} 3x \minus{} 2 \equal{} 0\] We can factor this as \[ (2x \plus{}1)(x \minus{} 2) \equal{} 0\] which gives us $ x \equal{} \minus{}\frac{1}{2}, 2$. Since we know $ x$ is negative, our answer is -1/2.

um, the answer is 2/3. we can simply see that 3/2 is 1.5

The number EXCEEDS its reciprocal by 3/2

ok this has happened to me before i think coolmath should just report


So would you say that Intermediate classes help you answer almost any problem on the AIME pertraining to that subject on the AIME? Because I am getting bowled over on some of the harder counting and algebra problems on the AIME and nearly all of the trig/complex numbers. Do you think these classes would remedy this if I put a lot of effort into them?

[quote="1234567890"]So would you say that Intermediate classes help you answer almost any problem on the AIME pertraining to that subject on the AIME?[/quote]Yes.  I can't speak to our Intermediate Algebra class (because I'm not as familiar with the material, having never personally taught it), but both the Intermediate Counting and the Trig/Complex class should prepare you for pretty much any problem you might encounter on the AIME.

Of course, this is not a guarantee that you'll be able to solve the problem, but you will at least have the tools to make a credible attempt at a solution.


$ (a^2 \plus{} a x \minus{} x^2) (a^2 \plus{} 2 a x \minus{} x^2) \equal{} 0$

Use quadratic equation on both terms to find the 4 different values of $ a$ in terms of $ x$ if $ x \ne 0$ and $ (a,x)\equal{}(0,0)$ if $ x\equal{}0$.


This problem is quite fun:

A circle is inscribed inside a right triangle with legs [i]a

[/i] and [i]b[/i] and hypotenuse [i]c[/i]. Prove that

the inradius (radius of the circle inside the triangle) is

 ([i]a[/i]+[i]b[/i]-[i]c[i])/2

Solution: [hide]Since each leg of the triangle is

tangent to the circle, each leg can be broken into two

different parts. Also, from Power of the Point

formulas, you know that two tangents from the same

point have equal lengths. Therefore, you can split [i]a

[/i] into x and y, split [i]b[/i] into y and z, and [i]c[/i]

into x and z. [i]a[/i] + [i]b[/i] would equal X+2Y+Z.

If you subtract [i]c[/i] from that, you would get

X+2Y+Z-X-Z, or 2Y. Now, you should realize that

since [i]abc[/i] is a right triangle and that whenever a

radius goes to a tangent, it forms a right angle. Also,

(obviously) Y=Y. Therefore, the quadrilateral formed

by X, Y, and two radii (near the intersection of [i]a[/i]

and [i]b[/i]) must be a square. Going back to the

original algebra, you have 2Y remaining, as well as

the -/2 part of the formula to prove. 2Y/2 = Y, so you

know that ([i]a[/i]+[i]b[/i]-[i]c[i])/2 is equal to Y.

Finally, since (from the geometry part) you have the

square formed by X and Y, you know that Y is equal

to the inradius.

Therefore, ([i]a[/i]+[i]b[/i]-[i]c[i])/2 must equal the

inradius for every right triangle with legs [i]a[/i], [i]b

[/i], and [i]c[/i].[/hide]

Good luck!

[quote="nat mc"]What is  (a+b-c[b][i][/b])/2?[/quote]

he accidentaly italicised twice. it should be (a+b-c)\2

ok I see.

hmm, I'm thinking about how to approch this, but I'm too tie and want to go to bed  :sleeping:

Why did you give us the problem and practically the only solution... which basically just involves a bunch of equal tangents and the fact that the inradius forms a square with the right angle.

I think the solution is if you can't get it >.> . Kinda like challenge questions.

[hide="solution"]
First we, draw the radii from the center, to the swides a and b.  Since these lines are to a tangent point, they both make a right angle.  Since it is a right triangle that means the angle between a and b is right, making the last angle between the two radii also right.

Since, the two radii are equal, this is a square.  So, the radius to b is the same length as the tangent point on a to the vertex between a and b.  Call it r.  It is the same on side b.  Now call, on a, the distance between the tangent point and the other radius, a-r, same on b, it is b-r.

 You have to kinda know some geometry that two tangents from a point outside the circle are equal in length.  That makes, on side c, the length to the vertice between a and c, a-r, and the other is b-r.  So a-r+b-r=c  r=(a+b-c)/2

That was a really long way of saying that a-r+b-r=c which you could easily see by drawing it.
[/hide]

let the center of circle be O. Join OA ,OB ,OC.area of triangle ABC ( let radius be x)
= (1/2 * Ax) + (1/2 * Bx) + (1/2*Cx) also = 1/2 *A*B
x common... x/2(A+B+C).1/2 AB

therefore , x=AB/ A+B+C

= AB(A+B-C)/ A+B+C(A+B-C)
=AB(A+B-C)/ a^2+2ab + b^2 - c^2
a^2+B^2= C^2
therefore X= AB(A+B-C)/2AB
= A+B-C/2

hence proved

i like algebra more because it seems easier to me.

Wow... I guess I like Algebra much more, because I don't want to do this proof at all. :huh:

[quote="Xantos C. Guin"]I like Geometry because it is challenging and requires you to think.  Algebra is too straightforward. $x^2-7x+12=0$- just factor and solve. Geometry problems have more creativity involved.

[quote="4everwise"]I don't think I like either algebra or geometry more than the other. I like a combination of the two.[/quote]

Are you reffering to Algeombra?[/quote]
Algebra does not only include quadtratic equations and straightforwardness.  There is a lot of algebra on for example AIME, where it pretty much gives you nothing and you have t solve by using algebra, that you have made from the info you are given

i like both algebra and geometry
I like geometry because one needs to be able to think a little abstractly about the problms; however, algebra is sometimes like that however, algebra is sometimes quite intuitive

algebra is less creative than geometry I think, more just proceeding carefully and adding bits of strange manipulation when you need it.  geometry is more fun, but I'm much better at algebra than I am at geometry.  geometry at olympiad levels is really really interestnig, lots of creative things happen and you really need to bust your brain.  not that im capable of olympiad geometry....

personally i prefer geometry but within geometry algebra can be incorporated, and that's what makes difficult geometry problems fun

I don't really like geometry that much even though I'm pretty good at it.  Algebra is funner because its much easier to prove things if you have to type like I' doing now

algebra is hard. to a degree
geometry is hard. to a degree
algebra is rewarding.
geometry is not so rewarding.
In algebra most of the time you don't know the answer and are finding it. This makes algebra rewarding.
In geometry most of the time you know the answer and need to find out why the answe is the way the answer is. any other geometry problems could be considered algebra, right? So I conclude geometry is not rewarding. You didn't find the answer yourself.
Therefore algebra is better (more fun) than geometry.

ok.  :blush:
whoa.. just noticed the date mark!


Hi everyone!

I have a problem I've tried to solve.. English is my second language, so pardon me if I don't get all the mathterms right :-)

I tried to work out a simple moneymanagement forumla for investing. I have two kind of measures i want to "combine", the probabililty of an event wil happen and the effect of the cumulative profit (but in this case, drawdown).. and the value I want out of it is the probability both theese will happen. I'll try to explain further:

Let say that in every investment, I risk 1% of my equity. If I hit a drawdown of let say 30%, i'll stop.
By always calculating the risk as a percentage of my equity, i it will take nr of trades:

0,7 = 0,99^x      >       x = log(0,8)/log(0,99) = 35,48

At the same time, I want to include the probability of the event will happens. The probability of a winning trade is 51,27%...

What I want to know is the probability of the whole event to happen, how big the risk sholud be (in this example it's 1%) to have e.g under 0,1% chance.. then 1% chance etc..

You guys understand my point?

Thanks in advance!

/Johan

I feel like I need a little more information to understand what you want.  I am repeating a gamble, which I win with probability $p$ (say $p = 51.27\%$).  If I keep losing, I give up after a certain amount of time (which you calculated to be between 35 and 36 gambles).  What isn't clear is what happens if you win: do you stop immediately?  Or (if I keep going) how much will I win?  (In this second case, it's worth noting that the Gambler's Ruin will kick in if you keep going long enough, provided your odds of winning are not too high (the threshold is at least 50%, but I'm not sure how big).)

JBL: I posted another version of the question in the same forum, because I got no answer at first. The link is http://www.artofproblemsolving.com/Forum/viewtopic.php?t=110103.

I had one answer, but can't make so much out of it.. maybe you can help me with the next step?

/Flirrrt


[quote="Svejk"]For a=8 b=1 c=1 the inequality doesn't hold[/quote]
I don't think so ! :D
Can anybody do it? :)

Hi,

I think that this inequality does not hold for all $a,b,c>0$. For example lets take the limit of the first part while a goes to infinity and $b,c$ remain unchanged.
Then $\sqrt{\frac{a}{b+c}}$ goes to infinity too while the other two quantities vanish. So the limit of the first part is infinity.

The example given by Svejk is also a counter example since it gives

$\sqrt{\frac{8}{1+1}}+\sqrt{\frac{1}{1+8}}+\sqrt{\frac{1}{8+1}}\simeq 2.667 > \frac{3 \sqrt{2}}{2}\simeq 2.121$

[quote="2006"]$\ a,b,c >0$ .prove that
$\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\leq\frac{3\sqrt{2}}{2}$.
   Good luck  :D[/quote]
This inequality is surely incorrect and you actually can't find an upper bound for the expretion you wrot because obviously you can take $a$ much larger than $b+c$ and $\sqrt{\frac{a}{b+c}}$ will be as large as you want.If I am not mistaken ther is a such kind of inequality,maybe you just made a typo writting the one above.Here is the inequality I am talking about
$\sqrt{\frac{a}{a+b}}+\sqrt{\frac{b}{b+c}}+\sqrt{\frac{c}{c+a}}\leq\frac{3\sqrt{2}}{2}$
In fact,this inequality has been dissused in the ML. :wink:

[quote="spx2"]Ticks if your argument is true than it could be applied to nesbitt inequality as well
(see nesbitt inequality here http://www.mathlinks.ro/Forum/viewtopic.php?highlight=nesbitt&t=42896)
but that is certainly true (altough i am a little confused about it).[/quote]

No... Nesbitt's Inequality is $\sum_{cyc}{\frac{a}{b+c}}\geq \frac{3}{2}$

[quote="spx2"]Ticks if your argument is true than it could be applied to nesbitt inequality as well
(see nesbitt inequality here http://www.mathlinks.ro/Forum/viewtopic.php?highlight=nesbitt&t=42896)
but that is certainly true (altough i am a little confused about it).[/quote]
You see,the Nesbitt's inequality is symetric but the inequality I mentioned is cyclic :wink: .


calculus
function
integration
symmetry
invariant
geometry
geometric transformation

The function $ f (x)$ is a complicated non-linear function. It satisfies, $ f (x) \plus{} f (1 \minus{} x) \equal{} 1$
Evaluate, $ \displaystyle\int_0^{1} f(x) dx$

P.S. Can anyone suggest me any good text book for learning these type of integrals? I mean a bit advanced things. I know the basics. What is the best textbook for learning advanced integrals?
Thanks in advance.

This one's a pure trick. You don't know anything except a symmetry relation- so you should try to exploit that symmetry by reflecting the function.

If a graph of $ y \equal{} f(x)$ has symmetrical transformation with respect to the point $ (a,\ b)$, find the equation of it?

Could anyone translate the above my context in plain English?

[hide="Answer"]\frac{1}{2}[/hide]

The constant function 1/2 satisfies such a condition. The wording of the problem suggests the answer is invariant of f.

I do know that, I gave a hint, please translate what I want to say into palin English?

[quote="kunny"]If a graph of $ y \equal{} f(x)$ has symmetrical transformation with respect to the point $ (a,\ b)$, find the equation of it?[/quote]

No one seemed to relpy my demand. I wouldn't  like to give the solution immediately, but...

Here is the solution of the problem.

$ f(x) \plus{} f(1 \minus{} x) \equal{} 1\Longleftrightarrow 2\cdot \frac {1}{2} \minus{} y \equal{} f\left(2\cdot \frac {1}{2} \minus{} x\right)$, which means that $ y \equal{} f(x)$ has symmetric with respect to the point

$ \left(\frac {1}{2},\ \frac {1}{2}\right)$, the equivalent translation of the area gives $ \int_0^1 f(x)\ dx \equal{} \frac {1}{2}$.

[quote="kunny"]...into [b]palin[/b] English[/quote]Does that include the accent? It sounds like a bad idea either way.

Ignoring the typo, we can't find $ f$. There are many such functions. The point is to show that all of them have the same integral, and here's where I would start:
What can you say about $ \int_0^1 f(x)\,dx$ and $ \int_0^1 f(1\minus{}x)\,dx$?

Bringing the $ y$-coordinate into it doesn't do much for me.

[quote="jmerry"][quote="kunny"]...into [b]palin[/b] English[/quote][/quote]

That was a typo, as you would guess, it should be [b]plain[/b] .

We are to find a right-angled triangle, aren' t you?

[quote="jmerry"][quote="kunny"]The point is to show that all of them have the same integral, and here's where I would start:
What can you say about $ \int_0^1 f(x)\,dx$ and $ \int_0^1 f(1 \minus{} x)\,dx$?

Bringing the $ y$-coordinate into it doesn't do much for me.[/quote][/quote] Would you please show complete solution? I am a nobo at integrals  :(

and what's about the text book? can you suggest me a good one?

Suppose that the point $ P(x,\ y)$ on the graph of $ y = f(x)$ is to translate symmetrically with respect to the point $ A(a,\ b)$
and the point $ P$ maps the point $ Q(X,\ Y)$, since $ A$ is the mid point of $ PQ$,
thus $ \frac {x + X}{2} = a,\ \frac {y + Y}{2} = b\Longleftrightarrow x = 2a - X, y = 2b - Y$.
$ \therefore (x,\ y)\in{y = f(x)}\Longleftrightarrow 2b - Y = f(2a - X)$, the equation after the translation is $ \boxed{2b - y = f(2a - x)}$.

Now,
[quote="kunny"]$ f(x) + f(1 - x) = 1\Longleftrightarrow 2\cdot \frac {1}{2} - y = f\left(2\cdot \frac {1}{2} - x\right)$, which means that $ y = f(x)$ has symmetric with respect to the point

$ \left(\frac {1}{2},\ \frac {1}{2}\right)$, the equivalent translation of the area gives $ \int_0^1 f(x)\ dx = \frac {1}{2}$.[/quote]

kunny

[quote="jmerry"][quote="kunny"]...into [b]palin[/b] English[/quote]Does that include the accent? It sounds like a bad idea either way.

[/quote]

*giggle* :trampoline:

You can't learn how to simplify integrals from a textbook. What you want is a problem book (and a good teacher).

Berkeley Problems in Mathematics is a good one  :D

[quote="Moonmathpi496"]The function $ f (x)$ is a complicated non-linear function. It satisfies, $ f (x) \plus{} f (1 \minus{} x) \equal{} 1$
Evaluate, $ \displaystyle\int_0^{1} f(x) dx$.[/quote]

Let $ 1 \minus{} x \equal{} t$, we have $ \int_0^1 f(1 \minus{} x)\ dx \equal{} \int_0^1 f(t)\ dt$

$ \therefore \int_0^1 f(x)\ dx \equal{} \int_0^1 \{1 \minus{} f(1 \minus{} x)\}\ dx \equal{} 1 \minus{} \int_0^1 f(x)\ dx$

$ \therefore \int_0^1 f(x)\ dx \equal{} \frac {1}{2}$.


topology
advanced fields
advanced fields unsolved

is there a sort of generalization of the mayer-vietoris sequence of de rham cohomology to arbitrary open coverings? in detail: let $ M$ be a smooth manifold and $ (U_i)$ an open covering of $ M$, then we have a canonical sequence

$ 0 \to \Omega^k(M) \to \prod_i \Omega^k(U_i) \to \prod_{i,j} \Omega^k(U_i \cap U_j) \to 0$

is it exact? clearly only the surjectivity at the end is of interest. I think the index set must be countable.

Bott and Tu use something similar in the proof that the de Rham cohomology is finite dimensional on a manifold. They don't explicitly construct that generalized sequence, but they induct in the canonical way. This leads me to believe that this generalization holds and is exact for countable indexing.

Note: You may need to refine the covering to a good covering (which can always be done) so that all nonempty finite intersections get you something diffeomorphic to $ \mathbb{R}^n$ and then use some sort of Poincare Lemma argument.

Hmm...I'll keep thinking about this.

In the form given above, this is clearly not true: take $ U_1\equal{}U_2\equal{}U_3\equal{}M$. But if you replace the third term by the corresponding cocycles, your claim is equivalent to $ \check H^1(U,\Omega^k)\equal{}0$ which is true since $ \Omega^k$ is a fine sheaf. owk


If a,b,c are real numbers such that

1) a+b+c>0

2) abc>0

3) ab+bc+ca>0

prove that a,b,c>0.

Wlog, we may assume that a  \leq b  \leq  c.
Suppose that a  \leq 0. Then, from (2), we have a < 0 and b < 0 and c > 0.
From (1), we also have c > -a-b > 0.

Thus ab+bc+ca = ab + c(a+b) < ab -(a+b) 2 = -a 2  - b 2  - ab, where ab > 0.
Then ab+bc+ca < 0, a contradiction.

Pierre.

Let f(x)=(x-a)(x-b)(x-c)=x 3 -px 2 +qx-r
which p=a+b+c,q=ab+bc+ca, and r=abc.
if one of a,b,c is <0 , assume it's z,then we have 0=z 3 -pz 2 +qz-r<0, contradiction.  (Each term is <0)
so a,b,c >0.

[quote="pbornsztein"]But, what if one is equal to 0? Does your your reasoning apply?

Pierre.[/quote]

if one of a,b,c is equal to 0 then abc=0, contradiction.


If $(a_n)_{n\ge 1}$ is an decreasing sequence of positive reals s.t. $\sum a_n$ diverges, show that $\sum \frac{a_n}{1+na_n}$ also diverges.

I remember Harazi posting a problem very similar to this one, which turned out to be wrong. I wonder if this was it.

First, $\frac{a_n}{1+na_n}$ is half of the harmonic mean of $a_n$ and $\frac1n$, so it is between half the minimum and the minimum of the two sequences. Now, we need to show that if $a_n$ is positive and decreasing and $\sum a_n$ diverges, $\sum \min(a_n,\frac1n)$ diverges.

For this, apply the condensation test: A series of decreasing positive terms $\sum x(n)$ converges if and only if $\sum 2^n\cdot x(2^n)$ converges.

In this case, the series we are interested in diverges if $\sum \min(2^na_{2^n},1)$ diverges. For that series to converge, all but finitely many terms must be $2^na_{2^n}$. Since the sum of these diverges by the condensation test, the whole series diverges and we are done.

Now if we had $\sum a_n$ and $\sum b_n$ divergent, we could construct a case in which $\sum\frac{a_nb_n}{a_n+b_n}$ converged.

Very nice. Another way would be noting that $\left(\frac{a_n}{1+na_n}\right)$ is also decreasing and positive, so we can directly apply the condensation test (that's what I did).

Yes. It's just that I've done the minimum before, and you're going to have to compare to it anyway. Any negative power mean is comparable to the minimum, and any positive power mean is comparable to the maximum.


When my pdf output file is created, the magnification is always set at 165% even if the Adobe default zoom preferences is set at 100%.  Can someone tell me how to ensure the pdf opens at 100% maginfication?

[quote="stevem"]Use the hyperref's package [i]pdfstartview[/i] to do this. For example
[code]\usepackage[pdfstartview={XYZ null null 1},bookmarks=false]{hyperref}[/code]Full details and loads of other options can be found in the [url=http://www.tug.org/applications/hyperref/manual.html]hyperref manual[/url].[/quote]

Perfect!  Many thanks.