Let $x, y$ be positive real numbers. If \[129-x^2=195-y^2=xy,\] then $x = \frac{m}{n}$ for relatively prime positive integers $m, n$. Find $100m+n$.

[i]Proposed by Michael Tang
1. We start with the given equations:
   \[
   129 - x^2 = xy \quad \text{and} \quad 195 - y^2 = xy
   \]
   From these, we can write:
   \[
   x^2 + xy = 129 \quad \text{and} \quad y^2 + xy = 195
   \]

2. Subtract the first equation from the second:
   \[
   y^2 + xy - (x^2 + xy) = 195 - 129
   \]
   Simplifying, we get:
   \[
   y^2 - x^2 = 66
   \]
   This can be factored using the difference of squares:
   \[
   (y - x)(y + x) = 66
   \]

3. We also know from the equations:
   \[
   x^2 + xy = 129
   \]
   and
   \[
   y^2 + xy = 195
   \]
   Adding these two equations:
   \[
   x^2 + xy + y^2 + xy = 129 + 195
   \]
   Simplifying, we get:
   \[
   x^2 + 2xy + y^2 = 324
   \]
   Recognizing the left-hand side as a perfect square:
   \[
   (x + y)^2 = 324
   \]
   Taking the square root of both sides:
   \[
   x + y = 18
   \]

4. Now we have two equations:
   \[
   (y - x)(y + x) = 66
   \]
   and
   \[
   y + x = 18
   \]
   Let \( y - x = a \) and \( y + x = 18 \). Then:
   \[
   a \cdot 18 = 66
   \]
   Solving for \( a \):
   \[
   a = \frac{66}{18} = \frac{11}{3}
   \]

5. We now have:
   \[
   y - x = \frac{11}{3}
   \]
   and
   \[
   y + x = 18
   \]
   Adding these two equations:
   \[
   (y - x) + (y + x) = \frac{11}{3} + 18
   \]
   Simplifying, we get:
   \[
   2y = \frac{11}{3} + \frac{54}{3} = \frac{65}{3}
   \]
   Solving for \( y \):
   \[
   y = \frac{65}{6}
   \]

6. Subtracting the two equations:
   \[
   (y + x) - (y - x) = 18 - \frac{11}{3}
   \]
   Simplifying, we get:
   \[
   2x = 18 - \frac{11}{3} = \frac{54}{3} - \frac{11}{3} = \frac{43}{3}
   \]
   Solving for \( x \):
   \[
   x = \frac{43}{6}
   \]

7. We need to find \( 100m + n \) where \( x = \frac{m}{n} \) and \( m \) and \( n \) are relatively prime. Here, \( x = \frac{43}{6} \), so \( m = 43 \) and \( n = 6 \).

8. Calculating \( 100m + n \):
   \[
   100 \cdot 43 + 6 = 4300 + 6 = 4306
   \]

The final answer is \(\boxed{4306}\)