Isosceles triangle $\triangle{ABC}$ has $\angle{ABC}=\angle{ACB}=72^\circ$ and $BC=1$. If the angle bisector of $\angle{ABC}$ meets $AC$ at $D$, what is the positive difference between the perimeters of $\triangle{ABD}$ and $\triangle{BCD}$?

[i]2019 CCA Math Bonanza Tiebreaker Round #2[/i]
1. Given that $\triangle{ABC}$ is an isosceles triangle with $\angle{ABC} = \angle{ACB} = 72^\circ$ and $BC = 1$. We need to find the positive difference between the perimeters of $\triangle{ABD}$ and $\triangle{BCD}$, where $D$ is the point where the angle bisector of $\angle{ABC}$ meets $AC$.

2. Let $AB = AC = x$. Since $\angle{ABC} = \angle{ACB} = 72^\circ$, $\angle{BAC} = 180^\circ - 2 \times 72^\circ = 36^\circ$.

3. The angle bisector of $\angle{ABC}$ splits $\angle{ABC}$ into two $36^\circ$ angles. Therefore, $\angle{ABD} = 36^\circ$ and $\angle{DBC} = 36^\circ$.

4. By the Angle Bisector Theorem, the ratio of the segments $AD$ and $DC$ is equal to the ratio of the sides $AB$ and $BC$. Since $AB = AC = x$ and $BC = 1$, we have:
   \[
   \frac{AD}{DC} = \frac{AB}{BC} = \frac{x}{1}
   \]
   Let $AD = k$ and $DC = \frac{k}{x}$.

5. Since $AD + DC = AC = x$, we have:
   \[
   k + \frac{k}{x} = x
   \]
   Solving for $k$, we get:
   \[
   k \left(1 + \frac{1}{x}\right) = x \implies k = \frac{x^2}{x+1}
   \]

6. Now, we calculate the perimeters of $\triangle{ABD}$ and $\triangle{BCD}$:
   - Perimeter of $\triangle{ABD}$:
     \[
     AB + BD + AD = x + BD + \frac{x^2}{x+1}
     \]
   - Perimeter of $\triangle{BCD}$:
     \[
     BC + BD + DC = 1 + BD + \frac{x^2}{x+1}
     \]

7. To find $BD$, we use the Law of Cosines in $\triangle{ABD}$ and $\triangle{BCD}$:
   - In $\triangle{ABD}$:
     \[
     BD^2 = AB^2 + AD^2 - 2 \cdot AB \cdot AD \cdot \cos(36^\circ)
     \]
     \[
     BD^2 = x^2 + \left(\frac{x^2}{x+1}\right)^2 - 2 \cdot x \cdot \frac{x^2}{x+1} \cdot \cos(36^\circ)
     \]
   - In $\triangle{BCD}$:
     \[
     BD^2 = BC^2 + DC^2 - 2 \cdot BC \cdot DC \cdot \cos(36^\circ)
     \]
     \[
     BD^2 = 1 + \left(\frac{x^2}{x+1}\right)^2 - 2 \cdot 1 \cdot \frac{x^2}{x+1} \cdot \cos(36^\circ)
     \]

8. Since $BD$ is common in both triangles, we can simplify the calculations by noting that the positive difference between the perimeters is:
   \[
   \left(x + BD + \frac{x^2}{x+1}\right) - \left(1 + BD + \frac{x^2}{x+1}\right) = x - 1
   \]

The final answer is $\boxed{1}$.