Input:
**User**:
Let $S$ be the set of points $(a,b)$ with $0 \le a,$ $b \le 1$ such that the equation
\[x^4 + ax^3 - bx^2 + ax + 1 = 0\]has at least one real root.  Determine the area of the graph of $S.$
Please reason step by step, and put your final answer within \boxed{}.

To determine the area of the set $ S $, we are given the condition that the quartic polynomial

$$
x^4 + ax^3 - bx^2 + ax + 1 = 0
$$

has at least one real root, for $ (a, b) \in [0,1] \times [0,1] $. Our goal is to find the area of the region in the unit square where this condition is satisfied.

---

### Step 1: Recognize the Polynomial Structure

The polynomial is symmetric in a special way. It is **palindromic**, meaning its coefficients read the same forwards and backwards. This allows us to use a substitution to simplify the equation.

Divide the equation by $ x^2 $ (valid for $ x \ne 0 $):

$$
x^2 + a x - b + \frac{a}{x} + \frac{1}{x^2} = 0
$$

Let $ y = x + \frac{1}{x} $. Then, we know:

$$
x^2 + \frac{1}{x^2} = y^2 - 2
$$

Substituting into the equation:

$$
y^2 - 2 + a y - b = 0 \quad \Rightarrow \quad y^2 + a y - (b + 2) = 0
$$

This is a quadratic in $ y $, and we are interested in whether this equation has a real solution $ y $ such that $ y = x + \frac{1}{x} $ for some real $ x \ne 0 $.

---

### Step 2: Determine the Range of $ y = x + \frac{1}{x} $

For real $ x \ne 0 $, the expression $ x + \frac{1}{x} $ has the range:

$$
(-\infty, -2] \cup [2, \infty)
$$

This is because:

- For $ x > 0 $, by AM-GM, $ x + \frac{1}{x} \ge 2 $
- For $ x < 0 $, $ x + \frac{1}{x} \le -2 $

So, the original equation has a real root if and only if the quadratic in $ y $ has at least one root in $ (-\infty, -2] \cup [2, \infty) $.

---

### Step 3: Analyze the Quadratic in $ y $

Let the quadratic be:

$$
f(y) = y^2 + a y - (b + 2)
$$

We want $ f(y) = 0 $ to have at least one root in $ (-\infty, -2] \cup [2, \infty) $.

Since the parabola opens upwards (coefficient of $ y^2 $ is positive), the condition for having a root in $ (-\infty, -2] \cup [2, \infty) $ is equivalent to:

- $ f(-2) \le 0 $ or $ f(2) \le 0 $

Let’s compute these:

- $ f(-2) = (-2)^2 + a(-2) - (b + 2) = 4 - 2a - b - 2 = 2 - 2a - b $
- $ f(2) = 2^2 + a(2) - (b + 2) = 4 + 2a - b - 2 = 2 + 2a - b $

So the conditions become:

- $ 2 - 2a - b \le 0 \Rightarrow 2a + b \ge 2 $
- $ 2 + 2a - b \le 0 \Rightarrow b \ge 2a + 2 $

Now, since $ a, b \in [0,1] $, the second inequality $ b \ge 2a + 2 $ is **never satisfied**, because the right-hand side is at least 2, while $ b \le 1 $. So the only valid condition is:

$$
2a + b \ge 2
$$

---

### Step 4: Determine the Region in the Unit Square

We now find the area of the region in the unit square $ [0,1] \times [0,1] $ where $ 2a + b \ge 2 $.

This is a linear inequality. The line $ 2a + b = 2 $ intersects the square at:

- When $ a = 0 $: $ b = 2 $ → outside the square
- When $ b = 0 $: $ a = 1 $
- When $ b = 1 $: $ 2a + 1 = 2 \Rightarrow a = 0.5 $

So the line passes through the points $ (1, 0) $ and $ (0.5, 1) $, and the region $ 2a + b \ge 2 $ lies **above** this line within the square.

This region is a **right triangle** with vertices at:

- $ (1, 0) $
- $ (1, 1) $
- $ (0.5, 1) $

The base of the triangle is from $ (1, 0) $ to $ (1, 1) $, which has length 1.

The height is from $ (1, 1) $ to $ (0.5, 1) $, which has length $ 0.5 $.

So the area is:

$$
\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 1 \times 0.5 = \frac{1}{4}
$$

---

### Final Answer

$$
\boxed{\dfrac{1}{4}}
$$

Solution:
Note that $x = 0$ cannot be a solution of the equation.  Dividing both sides by $x^2,$ we get
\[x^2 + ax - b + \frac{a}{x} + \frac{1}{x^2} = 0.\]Let $y = x + \frac{1}{x}.$  Then $x^2 - yx + 1 = 0.$  The discriminant of this quadratic is
\[y^2 - 4,\]so there is a real root in $x$ as long as $|y| \ge 2.$

Also, $y^2 = x^2 + 2 + \frac{1}{x^2},$ so
\[y^2 + ay - (b + 2) = 0.\]By the quadratic formula, the roots are
\[y = \frac{-a \pm \sqrt{a^2 + 4(b + 2)}}{2}.\]First, we notice that the discriminant $a^2 + 4(b + 2)$ is always positive.  Furthermore, there is a value $y$ such that $|y| \ge 2$ as long as
\[\frac{a + \sqrt{a^2 + 4(b + 2)}}{2} \ge 2.\]Then $a + \sqrt{a^2 + 4(b + 2)} \ge 4,$ or $\sqrt{a^2 + 4(b + 2)} \ge 4 - a.$  Both sides are nonnegative, so we can square both sides, to get
\[a^2 + 4(b + 2) \ge a^2 - 8a + 16.\]This simplifies to $2a + b \ge 2.$

[asy]
unitsize(3 cm);

fill((1/2,1)--(1,0)--(1,1)--cycle,gray(0.7));
draw((0,0)--(1,0)--(1,1)--(0,1)--cycle);
draw((1/2,1)--(1,0));

label("$0$", (0,0), S);
label("$1$", (1,0), S);
label("$a$", (1,0), E);
label("$0$", (0,0), W);
label("$1$", (0,1), W);
label("$b$", (0,1), N);
[/asy]

Thus, $S$ is the triangle whose vertices are $(1,0),$ $(1,1),$ and $\left( \frac{1}{2}, 1 \right),$ which has area $\boxed{\frac{1}{4}}.$


Input:
**User**:
An equiangular octagon has four sides of length $1$ and four sides of length $\frac{\sqrt{2}}{2}$, arranged so that no two consecutive sides have the same length. What is the area of the octagon?
Please reason step by step, and put your final answer within \boxed{}.

To find the area of the given equiangular octagon, we begin by understanding its structure and properties.

---

### **Understanding the Octagon**

An **equiangular octagon** has all internal angles equal. For an octagon, each internal angle is:

$$
\text{Internal angle} = \frac{(8 - 2) \cdot 180^\circ}{8} = 135^\circ
$$

This means that the **external angle** at each vertex is:

$$
\text{External angle} = 180^\circ - 135^\circ = 45^\circ
$$

Since the octagon is equiangular, the direction of each side changes by 45° from the previous one. The side lengths alternate between 1 and $\frac{\sqrt{2}}{2}$, and no two consecutive sides have the same length. This implies the sides alternate in the order:

$$
1, \frac{\sqrt{2}}{2}, 1, \frac{\sqrt{2}}{2}, \dots
$$

---

### **Modeling the Octagon as Vectors**

We can model the octagon as a closed polygon in the complex plane, where each side is a vector with a given length and direction. Starting from the origin, we can compute the coordinates of each vertex by summing the vectors.

Let’s define the direction of each side as follows:

- Side 1: $1$ at angle $0^\circ$
- Side 2: $\frac{\sqrt{2}}{2}$ at angle $45^\circ$
- Side 3: $1$ at angle $90^\circ$
- Side 4: $\frac{\sqrt{2}}{2}$ at angle $135^\circ$
- Side 5: $1$ at angle $180^\circ$
- Side 6: $\frac{\sqrt{2}}{2}$ at angle $225^\circ$
- Side 7: $1$ at angle $270^\circ$
- Side 8: $\frac{\sqrt{2}}{2}$ at angle $315^\circ$

We convert each side into a vector in the complex plane and sum them up. The result is zero, confirming that the octagon is closed.

---

### **Computing the Area Using the Shoelace Formula**

We compute the coordinates of the vertices by summing the vectors step by step, starting from the origin. The resulting coordinates are:

- $V_0 = (0, 0)$
- $V_1 = (1, 0)$
- $V_2 = \left(\frac{3}{2}, \frac{1}{2}\right)$
- $V_3 = \left(\frac{3}{2}, \frac{3}{2}\right)$
- $V_4 = (1, 2)$
- $V_5 = (0, 2)$
- $V_6 = \left(-\frac{1}{2}, \frac{3}{2}\right)$
- $V_7 = \left(-\frac{1}{2}, \frac{1}{2}\right)$
- $V_8 = (0, 0)$

Using the **shoelace formula**:

$$
\text{Area} = \frac{1}{2} \left| \sum_{i=0}^{7} (x_i y_{i+1} - x_{i+1} y_i) \right|
$$

We compute each term and sum them up:

$$
\text{Sum} = 0 + \frac{1}{2} + \frac{3}{2} + \frac{3}{2} + 2 + 1 + \frac{1}{2} + 0 = 7
$$

$$
\text{Area} = \frac{1}{2} \cdot 7 = \frac{7}{2}
$$

---

### **Conclusion**

After verifying the coordinates and applying the shoelace formula, we find that the area of the equiangular octagon with alternating side lengths of 1 and $\frac{\sqrt{2}}{2}$ is:

$$
\boxed{\dfrac{7}{2}}
$$

Solution:
The octagon can be partitioned into five squares and four half squares, each with side length $\sqrt{2}/2$, so its area is \[
\displaystyle\left(5+4 \cdot \frac{1}{2}\right)\left(\frac{\sqrt{2}}{2}\displaystyle\right)^{2}= \boxed{\frac{7}{2}}.
\][asy]
unitsize(2cm);
for (int i=0; i<4; ++i) {
for (int j=0; j<4; ++j) {
draw((i,0)--(i,3),dashed);
draw((0,j)--(3,j),dashed);
};}
draw((1,0)--(2,0)--(3,1)--(3,2)--(2,3)--(1,3)--(0,2)--(0,1)--cycle,linewidth(0.7));
for (int i=0; i<2; ++i) {
label("1",(0.5+2i,2.5),S);
label("1",(0.5+2i,0.5),N);}
label("$\frac{\sqrt{2}}{2}$",(0,1.5),E);
label("$\frac{\sqrt{2}}{2}$",(3,1.5),W);
[/asy]


Input:
**User**:
A sequence $(a_n)$ is defined as follows:
\[a_{i + 1} = \frac{1}{1 - a_i}\]for $i \ge 1.$  If $a_3 = a_1,$ compute $(a_9)^9.$
Please reason step by step, and put your final answer within \boxed{}.

To solve the problem, we are given a recursive sequence defined by:

$$
a_{i+1} = \frac{1}{1 - a_i}, \quad \text{for } i \geq 1
$$

and the condition that:

$$
a_3 = a_1
$$

We are asked to compute:

$$
(a_9)^9
$$

---

### Step 1: Express $ a_3 $ in terms of $ a_1 $

Using the recurrence relation:

- $ a_2 = \frac{1}{1 - a_1} $
- $ a_3 = \frac{1}{1 - a_2} = \frac{1}{1 - \frac{1}{1 - a_1}} $

Simplify the denominator:

$$
1 - \frac{1}{1 - a_1} = \frac{(1 - a_1) - 1}{1 - a_1} = \frac{-a_1}{1 - a_1}
$$

So:

$$
a_3 = \frac{1}{\frac{-a_1}{1 - a_1}} = \frac{1 - a_1}{-a_1} = \frac{a_1 - 1}{a_1}
$$

Given that $ a_3 = a_1 $, we equate:

$$
\frac{a_1 - 1}{a_1} = a_1
$$

Multiply both sides by $ a_1 $:

$$
a_1 - 1 = a_1^2
$$

Rearranging:

$$
a_1^2 - a_1 + 1 = 0
$$

---

### Step 2: Solve the quadratic equation

$$
a_1^2 - a_1 + 1 = 0
$$

Use the quadratic formula:

$$
a_1 = \frac{1 \pm \sqrt{(-1)^2 - 4(1)(1)}}{2} = \frac{1 \pm \sqrt{-3}}{2} = \frac{1 \pm i\sqrt{3}}{2}
$$

So $ a_1 $ is a complex number, and the sequence is defined over the complex numbers.

---

### Step 3: Determine the periodicity of the sequence

We now compute the next few terms to understand the behavior of the sequence.

- $ a_2 = \frac{1}{1 - a_1} $
- $ a_3 = \frac{1}{1 - a_2} = a_1 $ (given)
- $ a_4 = \frac{1}{1 - a_3} = \frac{1}{1 - a_1} = a_2 $
- $ a_5 = \frac{1}{1 - a_4} = \frac{1}{1 - a_2} = a_3 = a_1 $

So the sequence is **periodic with period 2**:

$$
a_1, a_2, a_1, a_2, a_1, a_2, \ldots
$$

Therefore, for any odd index $ n $, $ a_n = a_1 $, and for even $ n $, $ a_n = a_2 $.

Since 9 is odd, we have:

$$
a_9 = a_1
$$

---

### Step 4: Compute $ (a_9)^9 = (a_1)^9 $

We now compute $ a_1^9 $ using the relation $ a_1^2 = a_1 - 1 $, which comes from the quadratic equation.

Compute powers of $ a_1 $:

- $ a_1^2 = a_1 - 1 $
- $ a_1^3 = a_1 \cdot a_1^2 = a_1(a_1 - 1) = a_1^2 - a_1 = (a_1 - 1) - a_1 = -1 $
- $ a_1^4 = a_1 \cdot a_1^3 = a_1(-1) = -a_1 $
- $ a_1^5 = a_1 \cdot a_1^4 = a_1(-a_1) = -a_1^2 = -(a_1 - 1) = -a_1 + 1 $
- $ a_1^6 = a_1 \cdot a_1^5 = a_1(-a_1 + 1) = -a_1^2 + a_1 = -(a_1 - 1) + a_1 = -a_1 + 1 + a_1 = 1 $
- $ a_1^7 = a_1 \cdot a_1^6 = a_1 \cdot 1 = a_1 $
- $ a_1^8 = a_1 \cdot a_1^7 = a_1^2 = a_1 - 1 $
- $ a_1^9 = a_1 \cdot a_1^8 = a_1(a_1 - 1) = a_1^2 - a_1 = (a_1 - 1) - a_1 = -1 $

Thus:

$$
(a_9)^9 = (a_1)^9 = -1
$$

---

### Final Answer

$$
\boxed{-1}
$$

Solution:
First, if $a_3 = a_1,$ then
\[a_1 = a_3 = a_5 = a_7 = a_9,\]so $(a_9)^9 = (a_1)^9.$

We have that
\begin{align*}
a_2 &= \frac{1}{1 - a_1}, \\
a_3 &= \frac{1}{1 - a_2} = \frac{1}{1 - \frac{1}{1 - a_1}} = \frac{1 - a_1}{1 - a_1 - 1} = \frac{1 - a_1}{-a_1}.
\end{align*}Then
\[\frac{1 - a_1}{-a_1} = a_1,\]so $1 - a_1 = -a_1^2.$  Then $a_1^2 - a_1 + 1 = 0.$  Multiplying both sides by $a_1 + 1,$ we get
\[(a_1 + 1)(a_1 ^2 - a_1 + 1) = 0,\]so $a_1^3 + 1 = 0.$  Then $a_1^3 = -1,$ so $a_1^9 = (-1)^3 = \boxed{-1}.$