LiIIiam0216

We are asked to determine how many distinct paths can be followed to spell the word "MATH" starting from the origin \( M \), given that movements are only allowed up, down, left, and right.

The points corresponding to \( A \), \( T \), and \( H \) are labeled on the xy-plane, and each of these labels corresponds to a specific set of coordinates.

### Step 1: Understanding the Problem

The problem provides:

- \( M \) is at the origin, \( (0, 0) \).

- \( A \)'s are at \( (1,0) \), \( (-1,0) \), \( (0,1) \), and \( (0,-1) \).

- \( T \)'s are at \( (2,0) \), \( (1,1) \), \( (0,2) \), \( (-1,1) \), \( (-2,0) \), \( (-1,-1) \), \( (0,-2) \), and \( (1,-1) \).

- \( H \)'s are at \( (3,0) \), \( (2,1) \), \( (1,2) \), \( (0,3) \), \( (-1,2) \), \( (-2,1) \), \( (-3,0) \), \( (-2,-1) \), \( (-1,-2) \), \( (0,-3) \), \( (1,-2) \), and \( (2,-1) \).

We need to determine how many distinct paths can be followed to spell "MATH", moving from \( M \) to an \( A \), then from \( A \) to a \( T \), and finally from \( T \) to an \( H \).

### Step 2: Movement Considerations

We are allowed to move only up, down, left, and right. This restricts the possible movements between the points labeled \( M \), \( A \), \( T \), and \( H \).

- From \( M \) at \( (0, 0) \), we can move to any of the \( A \)'s at \( (1,0) \), \( (-1,0) \), \( (0,1) \), or \( (0,-1) \).

- From each \( A \), we can move to one of the \( T \)'s that are one unit away from the \( A \)'s.

- From each \( T \), we can move to one of the \( H \)'s that are one unit away from the \( T \)'s.

### Step 3: Counting the Distinct Paths

Let’s break down the path counting process step by step.

#### Paths from \( M \) to \( A \):

From \( M = (0, 0) \), there are 4 possible \( A \)'s:

- \( A_1 = (1, 0) \)

- \( A_2 = (-1, 0) \)

- \( A_3 = (0, 1) \)

- \( A_4 = (0, -1) \)

So, there are 4 choices for the first step.

#### Paths from \( A \) to \( T \):

From each \( A \), we can move to a neighboring \( T \) that is one unit away. Let's examine the options for each \( A \):

- From \( A_1 = (1, 0) \), the possible \( T \)'s are \( (2, 0) \), \( (1, 1) \), and \( (1, -1) \). This gives 3 choices.

- From \( A_2 = (-1, 0) \), the possible \( T \)'s are \( (-2, 0) \), \( (-1, 1) \), and \( (-1, -1) \). This gives 3 choices.

- From \( A_3 = (0, 1) \), the possible \( T \)'s are \( (0, 2) \), \( (1, 1) \), and \( (-1, 1) \). This gives 3 choices.

- From \( A_4 = (0, -1) \), the possible \( T \)'s are \( (0, -2) \), \( (1, -1) \), and \( (-1, -1) \). This gives 3 choices.

Thus, for each \( A \), there are 3 possible \( T \)'s, so the total number of ways to move from \( A \) to \( T \) is \( 3 \times 4 = 12 \).

#### Paths from \( T \) to \( H \):

From each \( T \), we can move to a neighboring \( H \) that is one unit away. There are 3 neighboring \( H \)'s for each \( T \) (similarly to the calculation above). Therefore, for each \( T \), there are 3 possible \( H \)'s.

Thus, for each \( T \), there are 3 possible \( H \)'s, so the total number of ways to move from \( T \) to \( H \) is \( 3 \times 12 = 36 \).

### Step 4: Total Number of Paths

Multiplying the number of choices at each step, we get:

\[
4 \times 3 \times 3 = 36.
\]

Thus, the total number of distinct paths that can be followed to spell the word "MATH" is \( \boxed{36} \).

The answer is BD = 15*sqrt(2) + 8.

The shortest altitude is 15.

The area is 16*sqrt(7).

To determine how many points \((x, y)\) where both \(x\) and \(y\) are positive integers lie below the hyperbola \(xy = 16\), we need to find the integer pairs \((x, y)\) such that \(xy < 16\).

### Step-by-Step Solution:

1. **Consider values of \(x\) and find corresponding \(y\) values**:

   For each \(x\), \(y\) must satisfy \(1 \leq y < \frac{16}{x}\).

2. **Calculate pairs for each \(x\)**:

   - \(x = 1\):

     \[
     xy < 16 \implies y < \frac{16}{1} \implies y < 16 \implies y = 1, 2, 3, \ldots, 15 \quad (\text{15 values})
     \]

   - \(x= 2\):
     \[
     xy < 16 \implies y < \frac{16}{2} \implies y < 8 \implies y = 1, 2, 3, \ldots, 7 \quad (\text{7 values})
     \]

   - \(x = 3\):
     \[
     xy < 16 \implies y < \frac{16}{3} \implies y < 5.33 \implies y = 1, 2, 3, 4, 5 \quad (\text{5 values})
     \]

   - \(x = 4\):
     \[
     xy < 16 \implies y < \frac{16}{4} \implies y < 4 \implies y = 1, 2, 3 \quad (\text{3 values})
     \]

   - \(x = 5\):
     \[
     xy < 16 \implies y < \frac{16}{5} \implies y < 3.2 \implies y = 1, 2, 3 \quad (\text{3 values})
     \]

   - \(x = 6\):
     \[
     xy < 16 \implies y < \frac{16}{6} \implies y < 2.67 \implies y = 1, 2 \quad (\text{2 values})
     \]

   - \(x = 7\):
     \[
     xy < 16 \implies y < \frac{16}{7} \implies y < 2.29 \implies y = 1, 2 \quad (\text{2 values})
     \]

   - \(x = 8\):
     \[
     xy < 16 \implies y < \frac{16}{8} \implies y < 2 \implies y = 1 \quad (\text{1 value})
     \]

   - \(x = 9\) and higher:
     \[
     xy < 16 \implies y < \frac{16}{x} \implies y < \frac{16}{x} \implies y = 1 \quad (\text{1 value if } x \leq 15 \text{ else no values})
     \]

3. **Count the total number of pairs**:

   Summing all the valid \(y\) values for each \(x\):
   \[
   15 + 7 + 5 + 3 + 3 + 2 + 2 + 1 + 1 = 39
   \]

Therefore, there are \( \boxed{39} \) points of the form \((x, y)\) where both coordinates are positive integers and lie below the hyperbola \(xy = 16\).

(a) Let w = a + bi and z = c + di.  The rest is expanding.

(b) Let w = a + bi and z = c + di.  Then

\[\dfrac{w + z}{1 + wz} = \dfrac{a + c + bi + di}{1 + (a + bi)(c + di)}\]

To express this in rectangular form, we can multiply the numerator and denominator by the conjugate:

\[\dfrac{a + c + bi + di}{1 + (a + bi)(c + di)} = \dfrac{(a + c + bi + di)((1 - (a + bi)(c + di))}{(1 + (a + bi)(c + di))(1 - (a + bi)(c + di))}\]

The denominator simplifies to (1 - (a^2 + b^2)(c^2 + d^2)), which is real.  The numerator simplifies to a^2 - b^2 + c^2 - d^2, which is also real.  Therefore, the complex number (z + w)/(zw + 1) is real.

The probability is 9/13.

The probability is 3/8.