## Monday, December 21, 2009

### Average Winning Roll

$n$ people are rolling a random real number between zero and one. The highest roll wins. What is the expected value of the winning roll?

Labels:
average,
calculus,
Combinatorics,
expected value,
Random,
roll

## Thursday, December 17, 2009

### Solution: Stones and Marbles

*Original problem: here*

*First, I apologize that the problem was not carefully worded. The problem statement should have been: find all initial conditions where it's always possible for Bob to empty all the three boxes.*

There are 3 boxes, and 2010 stones and 2010 marbles are put arbitrarily inside those three boxes.

At each step, Bob is allowed to either:

1. take a stone from one box, a marble from another box, and put them on the third box

2. add or subtract all boxes by the same number of stones

3. add or subtract all boxes by the same number of marbles

Prove or disprove, that by repeating these steps, Bob can empty all three boxes.

Labels:
change,
Combinatorics,
invariance,
modulo,
operation,
Solution,
span

## Wednesday, December 16, 2009

### Stones and Marbles

There are 3 boxes, and 2010 stones and 2010 marbles are put arbitrarily inside those three boxes.

At each step, Bob is allowed to either:

1. take a stone from one box, a marble from another box, and put them on the third box

2. add or subtract all boxes by the same number of stones

3. add or subtract all boxes by the same number of marbles

Prove or disprove, that by repeating these steps, Bob can empty all three boxes.

At each step, Bob is allowed to either:

1. take a stone from one box, a marble from another box, and put them on the third box

2. add or subtract all boxes by the same number of stones

3. add or subtract all boxes by the same number of marbles

Prove or disprove, that by repeating these steps, Bob can empty all three boxes.

Labels:
boxes,
Combinatorics,
marbles,
repetition,
steps,
stones

## Tuesday, December 15, 2009

### Twist And Spin

A matrix $M$ is skew-symmetric if and only if $M^{tr} = -M$. Let $S$ be the set of all 4x4 skew-symmetric matrices.

Suppose $F:\mathbb{R}^4 \to \mathbb{R}^4$ is a differentiable vector field in $\mathbb{R}^4$, we define the operation "twist" that maps each point in $\mathbb{R}^4$ to $S$ as follows:

If $F(x,y,z,w) = A\vec i + B \vec j + C \vec k + D \vec l$ then

$(\bold{twist} F)(x,y,z,w) = \left( \begin{array}{cccc}

0 & B_x - A_y & C_x - A_z & D_x - A_w \\

A_y-B_x & 0 & C_y - B_z & D_y - B_w \\

A_z-C_x & B_z - C_y & 0 & D_z - C_w \\

A_w - D_x & B_w - D_y & C_w - D_z & 0 \end{array} \right) $

Where $B_x$ denotes $\partial B / \partial x$ and so on. One way that makes it easier to memorize the formula is to associate the first column and row with $x$, second column and row with $y$, third with $z$, and last with $w$.

If $G : \mathbb{R}^4 \to S$ is a differentiable function that takes a point in $\mathbb{R}^4$ to a skew-symmetric matrix, we define the operation "spin" that produces a vector field in $\mathbb{R}^4$ as follows:

If $G(x,y,z,w) = \{ \sum_{i,j} G_{ij}(x,y,z,w) \}$ (where $G_{ij}$ is the entry at the $i$-th row and $j$-th column, and since $G$ is skew-symmetric, then $G_{i,i} = 0$ and $G_{ji} = -G_{ij}$)

Then

$(\bold{spin} G)(x,y,z,w) = (G_{23_w} + G_{34_y} + G_{42_z}) \vec i $

$+(G_{13_w} + G_{34_x} + G_{41_z}) \vec j$

$+(G_{12_w} + G_{24_x} + G_{41_y}) \vec k$

$+(G_{12_z} + G_{23_x} + G_{31_y}) \vec l$

where again $G_{ij_x}$ denotes $\partial G_{ij} / \partial x$ and so on.

Prove the following:

1. $(\bold{spin}(\bold{twist} F)) = \vec 0$

2. $\bigtriangledown \cdot ( \bold{spin} G) = 0$

3. If $F$ is a conservative vector field, then $\bold{twist} F = \vec 0$

## Twist

Suppose $F:\mathbb{R}^4 \to \mathbb{R}^4$ is a differentiable vector field in $\mathbb{R}^4$, we define the operation "twist" that maps each point in $\mathbb{R}^4$ to $S$ as follows:

If $F(x,y,z,w) = A\vec i + B \vec j + C \vec k + D \vec l$ then

$(\bold{twist} F)(x,y,z,w) = \left( \begin{array}{cccc}

0 & B_x - A_y & C_x - A_z & D_x - A_w \\

A_y-B_x & 0 & C_y - B_z & D_y - B_w \\

A_z-C_x & B_z - C_y & 0 & D_z - C_w \\

A_w - D_x & B_w - D_y & C_w - D_z & 0 \end{array} \right) $

Where $B_x$ denotes $\partial B / \partial x$ and so on. One way that makes it easier to memorize the formula is to associate the first column and row with $x$, second column and row with $y$, third with $z$, and last with $w$.

## Spin

If $G : \mathbb{R}^4 \to S$ is a differentiable function that takes a point in $\mathbb{R}^4$ to a skew-symmetric matrix, we define the operation "spin" that produces a vector field in $\mathbb{R}^4$ as follows:

If $G(x,y,z,w) = \{ \sum_{i,j} G_{ij}(x,y,z,w) \}$ (where $G_{ij}$ is the entry at the $i$-th row and $j$-th column, and since $G$ is skew-symmetric, then $G_{i,i} = 0$ and $G_{ji} = -G_{ij}$)

Then

$(\bold{spin} G)(x,y,z,w) = (G_{23_w} + G_{34_y} + G_{42_z}) \vec i $

$+(G_{13_w} + G_{34_x} + G_{41_z}) \vec j$

$+(G_{12_w} + G_{24_x} + G_{41_y}) \vec k$

$+(G_{12_z} + G_{23_x} + G_{31_y}) \vec l$

where again $G_{ij_x}$ denotes $\partial G_{ij} / \partial x$ and so on.

## Problems:

Prove the following:

1. $(\bold{spin}(\bold{twist} F)) = \vec 0$

2. $\bigtriangledown \cdot ( \bold{spin} G) = 0$

3. If $F$ is a conservative vector field, then $\bold{twist} F = \vec 0$

## Tuesday, December 8, 2009

### Dissecting a tetrahedron

Let $T_r$ denote a regular tetrahedron whose side length is $r$.

1. Prove that it is impossible to assemble eight $T_1$ into a $T_2$.

2. Prove that a tetrahedron can be cut into four optically congruent pieces. Two solids $A$ and $B$ are considered optically congruent if either $A$ or its mirror image can be rotated into $B$.

3. Prove that a $T_2$ can be cut into four $T_1$ and a regular octahedron whose side length is 1.

1. Prove that it is impossible to assemble eight $T_1$ into a $T_2$.

2. Prove that a tetrahedron can be cut into four optically congruent pieces. Two solids $A$ and $B$ are considered optically congruent if either $A$ or its mirror image can be rotated into $B$.

3. Prove that a $T_2$ can be cut into four $T_1$ and a regular octahedron whose side length is 1.

Labels:
3D,
Geometry,
optically congruent,
solid geometry,
tetrahedron

## Wednesday, December 2, 2009

### Solution: Inequality

Original problem: http://dharmath.thehendrata.com/2009/12/02/inequality/

For positive real number $a,b,c$ prove that

$2\sqrt{ab+bc+ca} \leq \sqrt{3} \sqrt[3]{(a+b)(b+c)(c+a)}$

For positive real number $a,b,c$ prove that

$2\sqrt{ab+bc+ca} \leq \sqrt{3} \sqrt[3]{(a+b)(b+c)(c+a)}$

### Inequality

For positive real number $a,b,c$ prove that

$2\sqrt{ab+bc+ca} \leq \sqrt{3} \sqrt[3]{(a+b)(b+c)(c+a)}$

solution: http://dharmath.thehendrata.com/2009/12/02/solution-inequality/

$2\sqrt{ab+bc+ca} \leq \sqrt{3} \sqrt[3]{(a+b)(b+c)(c+a)}$

solution: http://dharmath.thehendrata.com/2009/12/02/solution-inequality/

Labels:
homogeneous,
Inequality,
Solved,
symmetric

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