Let $O$, $A$, $B$, and $C$ be points in space such that $\angle AOB=60^{\circ}$, $\angle BOC=90^{\circ}$, and $\angle COA=120^{\circ}$. Let $\theta$ be the acute angle between planes $AOB$ and $AOC$. Given that $\cos^2\theta=\frac{m}{n}$ for relatively prime positive integers $m$ and $n$, compute $100m+n$.

[i]Proposed by Michael Ren[/i]
1. **Define the problem in terms of geometry and vectors:**
   - We are given four points \(O\), \(A\), \(B\), and \(C\) in space with specific angles between them: \(\angle AOB = 60^\circ\), \(\angle BOC = 90^\circ\), and \(\angle COA = 120^\circ\).
   - We need to find the acute angle \(\theta\) between the planes \(AOB\) and \(AOC\).

2. **Set up the coordinate system:**
   - Place \(O\) at the origin \((0,0,0)\).
   - Let \(A\), \(B\), and \(C\) be points on the unit sphere centered at \(O\) (i.e., \(OA = OB = OC = 1\)).

3. **Use the given angles to find the coordinates of \(A\), \(B\), and \(C\):**
   - Assume \(A = (1, 0, 0)\).
   - Since \(\angle AOB = 60^\circ\), \(B\) can be placed at \((\frac{1}{2}, \frac{\sqrt{3}}{2}, 0)\).
   - Since \(\angle BOC = 90^\circ\), \(C\) can be placed at \((0, 1, 0)\).

4. **Find the normal vectors to the planes \(AOB\) and \(AOC\):**
   - The normal vector to plane \(AOB\) is given by the cross product of vectors \(\overrightarrow{OA}\) and \(\overrightarrow{OB}\):
     \[
     \overrightarrow{OA} = (1, 0, 0), \quad \overrightarrow{OB} = \left(\frac{1}{2}, \frac{\sqrt{3}}{2}, 0\right)
     \]
     \[
     \overrightarrow{n_1} = \overrightarrow{OA} \times \overrightarrow{OB} = \begin{vmatrix}
     \mathbf{i} & \mathbf{j} & \mathbf{k} \\
     1 & 0 & 0 \\
     \frac{1}{2} & \frac{\sqrt{3}}{2} & 0
     \end{vmatrix} = (0, 0, \frac{\sqrt{3}}{2})
     \]
   - The normal vector to plane \(AOC\) is given by the cross product of vectors \(\overrightarrow{OA}\) and \(\overrightarrow{OC}\):
     \[
     \overrightarrow{OA} = (1, 0, 0), \quad \overrightarrow{OC} = (0, 1, 0)
     \]
     \[
     \overrightarrow{n_2} = \overrightarrow{OA} \times \overrightarrow{OC} = \begin{vmatrix}
     \mathbf{i} & \mathbf{j} & \mathbf{k} \\
     1 & 0 & 0 \\
     0 & 1 & 0
     \end{vmatrix} = (0, 0, 1)
     \]

5. **Find the cosine of the angle \(\theta\) between the normal vectors:**
   - The cosine of the angle between two vectors \(\overrightarrow{n_1}\) and \(\overrightarrow{n_2}\) is given by:
     \[
     \cos \theta = \frac{\overrightarrow{n_1} \cdot \overrightarrow{n_2}}{|\overrightarrow{n_1}| |\overrightarrow{n_2}|}
     \]
   - Calculate the dot product \(\overrightarrow{n_1} \cdot \overrightarrow{n_2}\):
     \[
     \overrightarrow{n_1} \cdot \overrightarrow{n_2} = (0, 0, \frac{\sqrt{3}}{2}) \cdot (0, 0, 1) = \frac{\sqrt{3}}{2}
     \]
   - Calculate the magnitudes of \(\overrightarrow{n_1}\) and \(\overrightarrow{n_2}\):
     \[
     |\overrightarrow{n_1}| = \sqrt{0^2 + 0^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \frac{\sqrt{3}}{2}, \quad |\overrightarrow{n_2}| = \sqrt{0^2 + 0^2 + 1^2} = 1
     \]
   - Therefore:
     \[
     \cos \theta = \frac{\frac{\sqrt{3}}{2}}{\frac{\sqrt{3}}{2} \cdot 1} = 1
     \]

6. **Find \(\cos^2 \theta\):**
   \[
   \cos^2 \theta = 1
   \]

7. **Express \(\cos^2 \theta\) in the form \(\frac{m}{n}\) and compute \(100m + n\):**
   - Here, \(\cos^2 \theta = 1 = \frac{1}{1}\), so \(m = 1\) and \(n = 1\).
   - Therefore, \(100m + n = 100 \cdot 1 + 1 = 101\).

The final answer is \(\boxed{101}\).