For every pair of reals $0 < a < b < 1$, we define sequences $\{x_n\}_{n \ge 0}$ and $\{y_n\}_{n \ge 0}$ by $x_0 = 0$, $y_0 = 1$, and for each integer $n \ge 1$: \begin{align*}
   x_n & = (1 - a) x_{n - 1} + a y_{n - 1},
   \\ y_n & = (1 - b) x_{n - 1} + b y_{n - 1}.
\end{align*} The [i]supermean[/i] of $a$ and $b$ is the limit of $\{x_n\}$ as $n$ approaches infinity. Over all pairs of real numbers $(p, q)$ satisfying $\left (p - \textstyle\frac{1}{2} \right)^2 + \left (q - \textstyle\frac{1}{2} \right)^2 \le \left(\textstyle\frac{1}{10}\right)^2$, the minimum possible value of the supermean of $p$ and $q$ can be expressed as $\textstyle\frac{m}{n}$ for relatively prime positive integers $m$ and $n$. Compute $100m + n$.

[i]Proposed by Lewis Chen[/i]
1. **Define sequences and initial conditions:**
   Given sequences \(\{x_n\}_{n \ge 0}\) and \(\{y_n\}_{n \ge 0}\) with initial conditions \(x_0 = 0\) and \(y_0 = 1\), and recurrence relations:
   \[
   x_n = (1 - a) x_{n - 1} + a y_{n - 1},
   \]
   \[
   y_n = (1 - b) x_{n - 1} + b y_{n - 1}.
   \]

2. **Define new sequences \(s_n\) and \(d_n\):**
   Let \(s_n = x_n + y_n\) and \(d_n = x_n - y_n\). Then, we have:
   \[
   d_n = x_n - y_n = (1 - a) x_{n - 1} + a y_{n - 1} - ((1 - b) x_{n - 1} + b y_{n - 1}).
   \]
   Simplifying, we get:
   \[
   d_n = (1 - a - (1 - b)) x_{n - 1} + (a - b) y_{n - 1} = (b - a) (x_{n - 1} - y_{n - 1}) = (b - a) d_{n - 1}.
   \]
   Since \(d_0 = x_0 - y_0 = 0 - 1 = -1\), we have:
   \[
   d_n = (b - a)^n \cdot d_0 = -(b - a)^n.
   \]

3. **Calculate \(s_n\):**
   \[
   s_n = x_n + y_n = (1 - a) x_{n - 1} + a y_{n - 1} + (1 - b) x_{n - 1} + b y_{n - 1}.
   \]
   Simplifying, we get:
   \[
   s_n = (2 - a - b) x_{n - 1} + (a + b) y_{n - 1} = 2 x_{n - 1}.
   \]

4. **Express \(x_n\) in terms of \(s_n\) and \(d_n\):**
   \[
   x_n = \frac{s_n + d_n}{2} = \frac{2 x_{n - 1} - (b - a)^n}{2}.
   \]
   Therefore:
   \[
   x_n = x_{n - 1} + \frac{a (b - a)^{n - 1}}{2}.
   \]

5. **Sum the series to find the limit:**
   Since \(x_0 = 0\), we have:
   \[
   x_n = \sum_{k = 0}^{n - 1} \frac{a (b - a)^k}{2}.
   \]
   This is a geometric series with sum:
   \[
   x_n = \frac{a}{2} \sum_{k = 0}^{n - 1} (b - a)^k = \frac{a}{2} \cdot \frac{1 - (b - a)^n}{1 - (b - a)}.
   \]
   As \(n \to \infty\), \((b - a)^n \to 0\) since \(0 < b - a < 1\). Thus:
   \[
   \lim_{n \to \infty} x_n = \frac{a}{2(1 - (b - a))} = \frac{a}{2 - a - b}.
   \]

6. **Minimize the supermean:**
   We need to minimize \(\frac{a}{2 - a - b}\) over all pairs \((p, q)\) satisfying \(\left(p - \frac{1}{2}\right)^2 + \left(q - \frac{1}{2}\right)^2 \le \left(\frac{1}{10}\right)^2\). This is equivalent to finding the minimum value of \(\frac{p}{2 - p - q}\).

7. **Geometric interpretation:**
   The condition describes a circle centered at \(\left(\frac{1}{2}, \frac{1}{2}\right)\) with radius \(\frac{1}{10}\). The minimum value of \(\frac{p}{2 - p - q}\) occurs when \(p\) and \(q\) are as small as possible within the circle.

8. **Find the minimum value:**
   The minimum value occurs at the point on the circle closest to the origin, which is \(\left(\frac{1}{2} - \frac{1}{10}, \frac{1}{2} - \frac{1}{10}\right) = \left(\frac{2}{5}, \frac{2}{5}\right)\). Thus:
   \[
   \frac{p}{2 - p - q} = \frac{\frac{2}{5}}{2 - \frac{2}{5} - \frac{2}{5}} = \frac{\frac{2}{5}}{\frac{6}{5}} = \frac{2}{6} = \frac{1}{3}.
   \]

The final answer is \( \boxed{ 100m + n = 100 \cdot 1 + 3 = 103 } \).