[u]Set 2[/u]


[b]p4.[/b] $\vartriangle ABC$ is an isosceles triangle with $AB = BC$. Additionally, there is $D$ on $BC$ with $AC = DA = BD = 1$. Find the perimeter of $\vartriangle ABC$.


[b]p5[/b]. Let $r$ be the positive solution to the equation $100r^2 + 2r - 1 = 0$. For an appropriate $A$, the infinite series $Ar + Ar^2 + Ar^3 + Ar^4 +...$ has sum $1$. Find $A$.


[b]p6.[/b] Let $N(k)$ denote the number of real solutions to the equation $x^4 -x^2 = k$. As $k$ ranges from $-\infty$  to $\infty$, the value of $N(k)$ changes only a finite number of times. Write the sequence of values of $N(k)$ as an ordered tuple (i.e. if $N(k)$ went from $1$ to $3$ to $2$, you would write $(1, 3, 2)$).


PS. You should use hide for answers.
1. We start with the given equation \( x^4 - x^2 = k \). To find the number of real solutions, we can make a substitution. Let \( y = x^2 \). Then the equation becomes:
   \[
   y^2 - y - k = 0
   \]
   This is a quadratic equation in \( y \).

2. We solve the quadratic equation using the quadratic formula:
   \[
   y = \frac{1 \pm \sqrt{1 + 4k}}{2}
   \]
   Here, we need to consider the discriminant \( 1 + 4k \) to determine the number of real solutions for \( y \).

3. We analyze the discriminant \( 1 + 4k \) for different values of \( k \):
   - **Case 1:** \( k < -\frac{1}{4} \)
     \[
     1 + 4k < 0 \implies \text{No real solutions for } y
     \]
     Since \( y = x^2 \), there are no real solutions for \( x \). Thus, \( N(k) = 0 \).

   - **Case 2:** \( k = -\frac{1}{4} \)
     \[
     1 + 4k = 0 \implies y = \frac{1 \pm 0}{2} = \frac{1}{2}
     \]
     Since \( y = \frac{1}{2} \) is a single value, and \( y = x^2 \), there are two real solutions for \( x \) (i.e., \( x = \pm \sqrt{\frac{1}{2}} \)). Thus, \( N(k) = 2 \).

   - **Case 3:** \( -\frac{1}{4} < k < 0 \)
     \[
     1 + 4k > 0 \implies y = \frac{1 \pm \sqrt{1 + 4k}}{2}
     \]
     Both values of \( y \) are positive, so each \( y \) gives two real solutions for \( x \) (i.e., \( x = \pm \sqrt{y} \)). Thus, there are four real solutions for \( x \). Hence, \( N(k) = 4 \).

   - **Case 4:** \( k = 0 \)
     \[
     1 + 4k = 1 \implies y = \frac{1 \pm 1}{2} = 1 \text{ or } 0
     \]
     For \( y = 1 \), \( x = \pm 1 \) (two solutions). For \( y = 0 \), \( x = 0 \) (one solution). Thus, there are three real solutions for \( x \). Hence, \( N(k) = 3 \).

   - **Case 5:** \( k > 0 \)
     \[
     1 + 4k > 0 \implies y = \frac{1 \pm \sqrt{1 + 4k}}{2}
     \]
     Both values of \( y \) are positive, so each \( y \) gives two real solutions for \( x \) (i.e., \( x = \pm \sqrt{y} \)). Thus, there are four real solutions for \( x \). Hence, \( N(k) = 4 \).

4. Summarizing the values of \( N(k) \) as \( k \) ranges from \( -\infty \) to \( \infty \):
   \[
   \boxed{(0, 2, 4, 2, 3)}
   \]