A point is chosen uniformly at random from the interior of a unit square. Let $p$ be the probability that any circle centered at the point that intersects a diagonal of the square must also intersect a side of the square. Given that $p^2$ can be written as $m-\sqrt{n}$ for positive integers $m$ and $n$, what is $m+n$?


[i]2021 CCA Math Bonanza Lightning Round #3.1[/i]
1. **Define the problem and setup the geometry:**
   - Consider a unit square \(ABCD\) with diagonals intersecting at point \(X\).
   - Let \(M\) be the midpoint of side \(AB\).
   - We need to find the probability \(p\) that any circle centered at a randomly chosen point \(O\) inside the square, which intersects a diagonal, also intersects a side of the square.

2. **Analyze the condition:**
   - The condition is that the distance from \(O\) to the nearest side of the square is less than the distance from \(O\) to the nearest diagonal.
   - Without loss of generality, assume \(O\) is in triangle \(AMX\).

3. **Use angle bisector theorem:**
   - Let \(T\) be the point where the angle bisector of \(\angle XAM\) intersects \(XM\).
   - The condition holds when \(O\) is in triangle \(AMT\).

4. **Calculate the area ratio:**
   - The probability \(p\) is given by the ratio of the area of \(\triangle AMT\) to the area of \(\triangle AMX\).
   - Using the angle bisector theorem, we have:
     \[
     \frac{[AMT]}{[AMX]} = \frac{MT}{MX}
     \]
   - Since \(T\) is on the angle bisector, we use the angle bisector theorem:
     \[
     \frac{MT}{MX} = \frac{1}{1 + \frac{AX}{AM}}
     \]
   - In a unit square, the length of the diagonal \(AX\) is \(\sqrt{2}\) and the length of \(AM\) is \(\frac{1}{2}\):
     \[
     \frac{AX}{AM} = \frac{\sqrt{2}}{\frac{1}{2}} = 2\sqrt{2}
     \]
   - Therefore:
     \[
     \frac{MT}{MX} = \frac{1}{1 + 2\sqrt{2}}
     \]

5. **Simplify the expression:**
   - Simplify the fraction:
     \[
     p = \frac{1}{1 + 2\sqrt{2}}
     \]
   - Rationalize the denominator:
     \[
     p = \frac{1}{1 + 2\sqrt{2}} \cdot \frac{1 - 2\sqrt{2}}{1 - 2\sqrt{2}} = \frac{1 - 2\sqrt{2}}{1 - (2\sqrt{2})^2} = \frac{1 - 2\sqrt{2}}{1 - 8} = \frac{1 - 2\sqrt{2}}{-7} = \frac{2\sqrt{2} - 1}{7}
     \]

6. **Calculate \(p^2\):**
   - Square \(p\):
     \[
     p^2 = \left(\frac{2\sqrt{2} - 1}{7}\right)^2 = \frac{(2\sqrt{2} - 1)^2}{49} = \frac{8 - 4\sqrt{2} + 1}{49} = \frac{9 - 4\sqrt{2}}{49}
     \]
   - Simplify:
     \[
     p^2 = \frac{9}{49} - \frac{4\sqrt{2}}{49} = \frac{9 - 4\sqrt{2}}{49}
     \]

7. **Identify \(m\) and \(n\):**
   - Given \(p^2 = m - \sqrt{n}\), we compare:
     \[
     \frac{9 - 4\sqrt{2}}{49} = m - \sqrt{n}
     \]
   - Therefore, \(m = \frac{9}{49}\) and \(\sqrt{n} = \frac{4\sqrt{2}}{49}\).

8. **Find \(m+n\):**
   - Since \(m\) and \(n\) must be integers, we multiply both sides by 49:
     \[
     49p^2 = 9 - 4\sqrt{2}
     \]
   - Thus, \(m = 9\) and \(n = 8\).

The final answer is \( \boxed{ m+n = 9 + 8 = 17 } \).