kittykat

We are tasked with finding all possible values of \( b \) that satisfy the following system of equations involving the floor functions:

\[
\lfloor a \rfloor \cdot b \cdot c = 3 \tag{1}
\]

\[
a \cdot \lfloor b \rfloor \cdot c = 4 \tag{2}
\]

\[
a \cdot b \cdot \lfloor c \rfloor = 5 \tag{3}
\]

where \( a \), \( b \), and \( c \) are positive real numbers. We will solve this step by step by exploring possible values for \( a \), \( b \), and \( c \).

### Solution By Steps

#### Step 1: Analyze the first equation

From Equation (1):

\[
\lfloor a \rfloor \cdot b \cdot c = 3
\]

The floor function \( \lfloor a \rfloor \) represents the greatest integer less than or equal to \( a \). Since \( a \) is positive, the possible values of \( \lfloor a \rfloor \) are integers.

We first assume \( \lfloor a \rfloor = 1 \), which is the smallest possible value because \( a > 0 \). Substituting this into Equation (1):

\[
1 \cdot b \cdot c = 3
\]

Thus, we have:

\[
b \cdot c = 3 \tag{4}
\]

#### Step 2: Analyze the second equation

Now, consider Equation (2):

\[
a \cdot \lfloor b \rfloor \cdot c = 4
\]

We already know \( b \cdot c = 3 \) from Equation (4). To satisfy this equation, we explore possible values of \( \lfloor b \rfloor \).

Let’s assume \( \lfloor b \rfloor = 1 \) first. Substituting into Equation (2):

\[
a \cdot 1 \cdot c = 4
\]

So,

\[
a \cdot c = 4 \tag{5}
\]

#### Step 3: Analyze the third equation

Now consider Equation (3):

\[
a \cdot b \cdot \lfloor c \rfloor = 5
\]

We already have \( b \cdot c = 3 \) and \( a \cdot c = 4 \). To satisfy this equation, we explore possible values of \( \lfloor c \rfloor \).

Let’s assume \( \lfloor c \rfloor = 1 \). Substituting into Equation (3):

\[
a \cdot b \cdot 1 = 5
\]

So:

\[
a \cdot b = 5 \tag{6}
\]

#### Step 4: Solve the system of equations

Now, we have three equations:

1. \( b \cdot c = 3 \)

2. \( a \cdot c = 4 \)

3. \( a \cdot b = 5 \)

We can solve this system step by step. First, solve for \( c \) from Equation (5):

\[
c = \frac{4}{a}
\]

Substitute this into Equation (4):

\[
b \cdot \frac{4}{a} = 3
\]

Simplifying:

\[
b = \frac{3a}{4}
\]

Now substitute this into Equation (6):

\[
a \cdot \frac{3a}{4} = 5
\]

Simplifying:

\[
\frac{3a^2}{4} = 5
\]

Multiply both sides by 4:

\[
3a^2 = 20
\]

Solve for \( a^2 \):

\[
a^2 = \frac{20}{3}
\]

So:

\[
a = \sqrt{\frac{20}{3}} = \frac{2\sqrt{15}}{3}
\]

#### Step 5: Find the value of \( b \)

Now that we have \( a = \frac{2\sqrt{15}}{3} \), substitute this back into the expression for \( b \):

\[
b = \frac{3a}{4} = \frac{3 \times \frac{2\sqrt{15}}{3}}{4} = \frac{2\sqrt{15}}{4} = \frac{\sqrt{15}}{2}
\]

Thus, the value of \( b \) is:

\[
b = \frac{\sqrt{15}}{2}
\]

To determine the area of a square inscribed in a right triangle with legs of length 1 and 3, we can use a geometric approach.

Let's denote the right triangle \(ABC\) where \(A = (0, 0)\), \(B = (1, 0)\), and \(C = (0, 3)\). The right angle is at \(A\), the x-axis extends along \(AB\), and the y-axis along \(AC\).

Let the side length of the inscribed square be \(s\). When a square is inscribed in the triangle, the square's lower left corner will coincide with point \(A\) (the right angle), and the top right corner will lie somewhere along the hypotenuse \(BC\).

We need to find the relationship between \(s\) and the coordinates of the square touching the hypotenuse. The coordinates where the square touches the hypotenuse will be at the point \((s, s)\) since the square's sides are parallel to the axes.

Now, we need to find the equation of line \(BC\). The slope \(m\) of line segment \(BC\) can be calculated as follows:

\[
m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - 0}{0 - 1} = -3.
\]

The equation of line \(BC\), which has a y-intercept at point \(C\) \((0, 3)\), can be expressed as:

\[
y - 3 = -3(x - 0) \quad \Rightarrow \quad y = -3x + 3.
\]

Now, we want the point \((s, s)\) on the line \(BC\). Therefore, substituting \(x = s\) into the line equation gives:

\[
s = -3s + 3.
\]

Solving this equation for \(s\):

\[
s + 3s = 3 \quad \Rightarrow \quad 4s = 3 \quad \Rightarrow \quad s = \frac{3}{4}.
\]

The area \(A\) of the square is given by \(s^2\):

\[
A = s^2 = \left( \frac{3}{4} \right)^2 = \frac{9}{16}.
\]

Thus, the area of the square inscribed in the right triangle is

\[
\boxed{\frac{9}{16}}.
\]

To determine which integers \( n \) in the range \( 70 \leq n \leq 90 \) can be expressed in the form

\[
n = ab + 2a + 3b,
\]

we can rearrange this equation:

\[
n = ab + 2a + 3b = ab + 2a + 3b = a(b + 2) + 3b.
\]

Thus, we can factor the expression:

\[
n = a(b + 2) + 3b = (b + 2)a + 3b.
\]

Rearranging further, we can consider the expression more carefully:

\[
n = ab + 2a + 3b = a(b + 2) + 3b.
\]

To find how \( n \) varies based on \( a \) and \( b \), we can think of expressions:

\[
n = ab + 2a + 3b = a(b + 2) + 3b.
\]

To explore values of \( n \), we can set specific values for \( b \) and investigate how \( n \) changes. We start by expressing \( a \) in terms of \( n \) for different values of \( b \):

1. Rearranging gives us:

\[
n - 3b = a(b + 2).
\]

This implies

\[
a = \frac{n - 3b}{b + 2}.
\]

For \( a \) to be a positive integer, \( n - 3b \) must be divisible by \( b + 2 \) and \( n - 3b > 0 \). This leads to key constraints:

- \( n > 3b \) (or equivalently, \( b < \frac{n}{3} \)),

- \( n - 3b \) should be divisible by \( b + 2 \).

Now, let’s analyze integers within the specified range \( 70 \leq n \leq 90 \).

**For \( b = 1 \):**

\[
n = a(1 + 2) + 3 \cdot 1 = 3a + 3 \Rightarrow n - 3 = 3a \Rightarrow n = 3a + 3 \Rightarrow n - 3 \equiv 0 \, (\text{mod } 3).
\]

Thus, \( n \equiv 0 \, (\text{mod } 3) \) which gives us possible \( n \): \( 72, 75, 78, 81, 84, 87, 90 \).

**For \( b = 2 \):**

\[
n = a(2 + 2) + 3 \cdot 2 = 4a + 6 \Rightarrow n - 6 = 4a \Rightarrow n = 4a + 6 \Rightarrow n - 6 \equiv 0 \, (\text{mod } 4).
\]

Thus, \( n \equiv 2 \, (\text{mod } 4) \) which gives us \( n \): \( 70, 74, 78, 82, 86, 90 \).

**For \( b = 3 \):**

\[
n = a(3 + 2) + 3 \cdot 3 = 5a + 9 \Rightarrow n - 9 = 5a \Rightarrow n = 5a + 9 \Rightarrow n - 9 \equiv 0 \, (\text{mod } 5).
\]

Thus, \( n \equiv 4 \, (\text{mod } 5) \) which gives us \( n \): \( 74, 79, 84, 89 \).

**For \( b = 4 \):**

\[
n = a(4 + 2) + 3 \cdot 4 = 6a + 12 \Rightarrow n - 12 = 6a \Rightarrow n = 6a + 12 \Rightarrow n - 12 \equiv 0 \, (\text{mod } 6).
\]

Thus, \( n \equiv 0 \, (\text{mod } 6) \) which gives us \( n \): \( 72, 78, 84, 90 \).

Continuing this process for \( b = 5 \) and \( b = 6 \) eventually provides specific integers. We can compile our results and find the integers \( n \):

By synthesizing:

- All candidate results from \( b = 1, 2, 3, 4, 5, 6 \):

The integers derived include:

- \( n = 70, 72, 74, 75, 78, 79, 81, 82, 84, 86, 87, 89, 90 \).

Finally, we count unique integers:

\[
\{70, 72, 74, 75, 78, 79, 81, 82, 84, 86, 87, 89, 90\}
\]

Counting these gives:

\[
\text{Total count} = 13.
\]

Thus, the answer is

\[
\boxed{13}.
\]

We want to find the exponential form of the complex number:

\[
e^{\frac{17\pi i}{60}} + e^{\frac{27\pi i}{60}} + e^{\frac{37\pi i}{60}} + e^{\frac{47\pi i}{60}} + e^{\frac{57\pi i}{60}}
\]

### Step 1: Recognize the terms as roots of unity

The terms \( e^{\frac{17\pi i}{60}}, e^{\frac{27\pi i}{60}}, e^{\frac{37\pi i}{60}}, e^{\frac{47\pi i}{60}}, e^{\frac{57\pi i}{60}} \) are all complex numbers in exponential form. Notice that these exponents can be expressed as:

\[
\frac{17\pi i}{60}, \frac{27\pi i}{60}, \frac{37\pi i}{60}, \frac{47\pi i}{60}, \frac{57\pi i}{60}
\]

These correspond to angles \( \theta = \frac{17\pi}{60}, \frac{27\pi}{60}, \frac{37\pi}{60}, \frac{47\pi}{60}, \frac{57\pi}{60} \).

Since \( e^{i\theta} \) represents a point on the unit circle in the complex plane, each of these terms can be considered as specific roots of unity, although not all are primitive roots.

### Step 2: Consider the sum of the angles

The sum can be expressed as:

\[
S = e^{\frac{17\pi i}{60}} + e^{\frac{27\pi i}{60}} + e^{\frac{37\pi i}{60}} + e^{\frac{47\pi i}{60}} + e^{\frac{57\pi i}{60}}
\]

These angles are evenly spaced on the unit circle by an angle increment of \( \frac{10\pi}{60} = \frac{\pi}{6} \).

### Step 3: Utilize symmetry

These angles correspond to the roots of the equation \( x^5 - 1 = 0 \) rotated by a small angle \( \frac{17\pi}{60} \). The angles \( \frac{17\pi}{60}, \frac{27\pi}{60}, \frac{37\pi}{60}, \frac{47\pi}{60}, \frac{57\pi}{60} \) map to the five vertices of a regular pentagon inscribed in the unit circle but rotated slightly.

For a regular \( n \)-gon (in this case, \( n = 5 \)), the sum of vectors corresponding to the vertices is zero if the vectors are evenly distributed around the circle (because they symmetrically cancel each other out). However, here, the vertices are slightly rotated, but they are still symmetrically spaced around the circle.

### Step 4: Apply properties of roots of unity

Given the symmetry and spacing:

\[
S = e^{i\frac{\theta_0}{60}} \cdot \left(1 + e^{i\frac{2\pi}{5}} + e^{i\frac{4\pi}{5}} + e^{i\frac{6\pi}{5}} + e^{i\frac{8\pi}{5}}\right)
\]

Where \( \theta_0 \) is \( \frac{17\pi}{60} \), and the roots sum to zero because they represent the vertices of a pentagon. Thus, the sum

:

\[
S = 0
\]

### Conclusion:

The exponential form of the given complex number is:

\[
\boxed{0}
\]

Given that triangle \(ABC\) has altitudes \(AD = 14\), \(BE = 16\), and \(CF = h\), where \(h\) is a positive integer, we need to find the largest possible value of \(h\).

### Step 1: Use the area formula with altitudes

The area of triangle \(ABC\) can be expressed in terms of the altitudes and the corresponding sides:

\[
\text{Area} = \frac{1}{2} \times a \times h_a = \frac{1}{2} \times b \times h_b = \frac{1}{2} \times c \times h_c
\]

where \(h_a = AD\), \(h_b = BE\), and \(h_c = CF = h\), and \(a\), \(b\), and \(c\) are the lengths of the sides opposite the vertices \(A\), \(B\), and \(C\), respectively.

### Step 2: Express the sides in terms of the area and altitudes

Let the area of triangle \(ABC\) be denoted as \(K\). Then:

\[
K = \frac{1}{2} \times a \times 14 = 7a
\]

\[
K = \frac{1}{2} \times b \times 16 = 8b
\]

\[
K = \frac{1}{2} \times c \times h = \frac{1}{2}ch
\]

### Step 3: Equate the expressions for the area \(K\)

From the first two equations:

\[
7a = 8b \quad \Rightarrow \quad a = \frac{8b}{7}
\]

Using the equation for \(K\) involving \(h\):

\[
7a = \frac{1}{2} ch
\]

Substitute \(a = \frac{8b}{7}\) into the equation:

\[
7\left(\frac{8b}{7}\right) = \frac{1}{2}ch
\]

Simplify:

\[
8b = \frac{1}{2}ch
\]

Multiply both sides by 2:

\[
16b = ch
\]

Thus:

\[
h = \frac{16b}{c}
\]

### Step 4: Maximize \(h\)

We aim to maximize \(h = \frac{16b}{c}\), where \(b\) and \(c\) are positive integers.

Recall that \(K = 7a = 8b = \frac{1}{2}ch\). We want to maximize \(h\), so:

\[
c = \frac{16b}{h}
\]

To maximize \(h\), minimize \(c\). Since \(b = 7\), \(a = 8\), choose \(b = 7\) and \(c = 1\) which gives:

\[
h = \frac{16 \times 7}{1} = 112
\]

However, since \(b = \gcd(7,8)\), choose \(b = 7\) for simplicity. So:

\[
K = 112, h = 1
]

Thus:

\[
h \leq 112
\]

### Final Answer:

Since \(h = 112\), the maximum value for the given maximum number is \( \boxed{112} \).

To determine the number of points of the form \((x,y)\), where both coordinates are positive integers, that lie below the graph of the hyperbola \(xy = 16\), we need to find the integer pairs \((x, y)\) such that \(xy < 16\).

### Step-by-Step Solution:

1. **Identify the Range for \(x\)**:

   - For each \(x\), we need \(y\) to be an integer such that \(xy < 16\).

   - Since \(x\) must be a positive integer, consider possible values of \(x\) starting from 1 up to the point where \(x \cdot 1 = 16\), so \(x\) ranges from 1 to 15.

2. **Count \(y\) Values for Each \(x\)**:

   - For each \(x\), find the largest integer \(y\) such that \(y < \frac{16}{x}\).

   Here’s how this works for each \(x\) from 1 to 15:

   - \(x = 1\): \(xy < 16 \implies y < \frac{16}{1} = 16\). So, \(y\) can be 1 to 15 (15 values).

   - \(x = 2\): \(xy < 16 \implies y < \frac{16}{2} = 8\). So, \(y\) can be 1 to 7 (7 values).

   - \(x = 3\): \(xy < 16 \implies y < \frac{16}{3} \approx 5.33\). So, \(y\) can be 1 to 5 (5 values).

   - \(x = 4\): \(xy < 16 \implies y < \frac{16}{4} = 4\). So, \(y\) can be 1 to 3 (3 values).

   - \(x = 5\): \(xy < 16 \implies y < \frac{16}{5} = 3.2\). So, \(y\) can be 1 to 3 (3 values).

   - \(x = 6\): \(xy < 16 \implies y < \frac{16}{6} \approx 2.67\). So, \(y\) can be 1 to 2 (2 values).

   - \(x = 7\): \(xy < 16 \implies y < \frac{16}{7} \approx 2.29\). So, \(y\) can be 1 to 2 (2 values).

   - \(x = 8\): \(xy < 16 \implies y < \frac{16}{8} = 2\). So, \(y\) can be 1 (1 value).

   - \(x = 9\) to \(x = 15\): For these values, \(y < \frac{16}{x}\) will always be less than 2, so \(y\) can only be 1 (1 value each).

3. **Summarize the Counts**:

   \[
   \begin{aligned}

   &15 \text{ values for } x = 1, \\

   &7 \text{ values for } x = 2, \\

   &5 \text{ values for } x = 3, \\

   &3 \text{ values for } x = 4, \\

   &3 \text{ values for } x = 5, \\

   &2 \text{ values for } x = 6, \\

   &2 \text{ values for } x = 7, \\

   &1 \text{ value for } x = 8, \\

   &1 \text{ value each for } x = 9 \text{ to } 15.

   \end{aligned}

   \]

   Adding these up:

   \[
   15 + 7 + 5 + 3 + 3 + 2 + 2 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 44
   \]

Therefore, the number of points \((x, y)\), where both coordinates are positive integers, that lie below the graph of the hyperbola \(xy = 16\) is:
\[
\boxed{44}
\]