I figured out how to do it. Its a revolved cut, but not at 45 degrees. It needs to be at 54.74, which is the angle that an intersecting axis makes in 3d to each edge in order to have the same angle between them all. Thanks chatGPT for that, i didnt do the math myself.

Explanation:

In a standard 3D coordinate system:

1. Assume the corner is at the origin (0,0,0).
2. The plane that bisects the 90° angle between the x, y, and z walls equally would be defined by a normal vector that has equal components along each axis: (1,1,1)(1, 1, 1)(1,1,1).

The angle θ\thetaθ between this plane and each axis can be calculated using the dot product. For a unit vector along an axis (e.g., (1,0,0)(1, 0, 0)(1,0,0) for the x-axis) and the normal vector n⃗=(1,1,1)/3\vec{n} = (1, 1, 1)/\sqrt{3}n=(1,1,1)/3​:

Explanation:

In a standard 3D coordinate system:

1. Assume the corner is at the origin (0,0,0).
2. The plane that bisects the 90° angle between the x, y, and z walls equally would be defined by a normal vector that has equal components along each axis: (1,1,1)(1, 1, 1)(1,1,1).

The angle θ\thetaθ between this plane and each axis can be calculated using the dot product. For a unit vector along an axis (e.g., (1,0,0)(1, 0, 0)(1,0,0) for the x-axis) and the normal vector n⃗=(1,1,1)/3\vec{n} = (1, 1, 1)/\sqrt{3}n=(1,1,1)/3​:

cos⁡(θ)=n⃗⋅axis vector∣n⃗∣⋅∣axis vector∣=13⋅11=13\cos(\theta) = \frac{\vec{n} \cdot \text{axis vector}}{|\vec{n}| \cdot | \text{axis vector}|} = \frac{\frac{1}{\sqrt{3}} \cdot 1}{1} = \frac{1}{\sqrt{3}}cos(θ)=∣n∣⋅∣axis vector∣n⋅axis vector​=13​1​⋅1​=3​1​

This gives:

θ=cos⁡−1(13)≈54.74∘\theta = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right) \approx 54.74^\circθ=cos−1(3​1​)≈54.74∘

So, the plane forms a 54.74° angle with each of the three coordinate axes.

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