Snub Cube Construction

Geometry Level 5

The so-called snub cube can be constructed from the regular cube as shown in the animation above. For each of the six faces, construct a directed axis passing through the face center and pointing outward from the cube. Now translate each of the six faces in the direction of the outward pointing axis by a distance d d . Next rotate all of the faces by a specific angle θ \theta either clockwise or counter clockwise about the directed axis. Note that the direction of rotation chosen has to apply to all the faces. Finally, connect all the vertices. The distance d d and the angle θ \theta are such that all the edges of the resulting solid are of the same length . If the side length of the original cube is 1 1 , find the displacement distance d d of each of the six faces of the original cube.


The answer is 0.6426.

Hosam Hajjir by Hosam Hajjir , 56, Canada

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1 solution

David Vreken
Oct 21, 2019

Let the coordinates of the unit cube be ( ± 1 2 , ± 1 2 , ± 1 2 ) (\pm \frac{1}{2}, \pm \frac{1}{2}, \pm \frac{1}{2}) , and specifically let A A be ( − 1 2 , 1 2 , 1 2 ) (-\frac{1}{2}, \frac{1}{2}, \frac{1}{2}) and B B be ( 1 2 , 1 2 , 1 2 ) (\frac{1}{2}, \frac{1}{2}, \frac{1}{2}) . Let A ′ A' be the new position of A A after a translation of d d in the positive y y direction and counter clockwise rotation of angle θ \theta about the y y -axis, B ′ B' be the new position of B B after a translation of d d in the positive y y direction and counter clockwise rotation of angle θ \theta about the y y -axis, and A ′ ′ A'' be the new position of A A after a translation of d d in the positive z z direction and counter clockwise rotation of angle θ \theta about the z z -axis, as shown below.

Then the new coordinates are:

A ′ ( − 1 2 cos ⁡ θ − 1 2 sin ⁡ θ , d + 1 2 , − 1 2 sin ⁡ θ + 1 2 cos ⁡ θ ) A'(-\frac{1}{2} \cos \theta - \frac{1}{2} \sin \theta, d + \frac{1}{2}, -\frac{1}{2} \sin \theta + \frac{1}{2} \cos \theta)

B ′ ( 1 2 cos ⁡ θ − 1 2 sin ⁡ θ , d + 1 2 , 1 2 sin ⁡ θ + 1 2 cos ⁡ θ ) B'(\frac{1}{2} \cos \theta - \frac{1}{2} \sin \theta, d + \frac{1}{2}, \frac{1}{2} \sin \theta + \frac{1}{2} \cos \theta)

A ′ ′ ( − 1 2 cos ⁡ θ + 1 2 sin ⁡ θ , 1 2 sin ⁡ θ + 1 2 cos ⁡ θ , d + 1 2 ) A''(-\frac{1}{2} \cos \theta + \frac{1}{2} \sin \theta, \frac{1}{2} \sin \theta + \frac{1}{2} \cos \theta, d + \frac{1}{2})

Since △ A ′ B ′ A ′ ′ \triangle A'B'A'' must be equilateral and A ′ B ′ = 1 A'B' = 1 ,

A ′ A ′ ′ = 1 = sin ⁡ 2 θ + ( 1 2 sin ⁡ θ + 1 2 cos ⁡ θ − d − 1 2 ) 2 + ( 1 2 sin ⁡ θ − 1 2 cos ⁡ θ + d + 1 2 ) 2 A'A'' = 1 = \sqrt{\sin^2 \theta + (\frac{1}{2} \sin \theta + \frac{1}{2} \cos \theta - d - \frac{1}{2})^2 + (\frac{1}{2} \sin \theta - \frac{1}{2} \cos \theta + d + \frac{1}{2})^2}

B ′ A ′ ′ = 1 = ( sin ⁡ θ − cos ⁡ θ ) 2 + ( 1 2 sin ⁡ θ + 1 2 cos ⁡ θ − d − 1 2 ) 2 + ( 1 2 sin ⁡ θ + 1 2 cos ⁡ θ − d − 1 2 ) 2 B'A'' = 1 = \sqrt{(\sin \theta - \cos \theta)^2 + (\frac{1}{2} \sin \theta + \frac{1}{2} \cos \theta - d - \frac{1}{2})^2 + (\frac{1}{2} \sin \theta + \frac{1}{2} \cos \theta - d - \frac{1}{2})^2}

Using wolframalpha.com these two equations solve numerically to θ ≈ 0.2874 \theta \approx 0.2874 and d ≈ 0.6426 d \approx \boxed{0.6426} .

1 Helpful 0 Interesting 1 Brilliant 0 Confused

That's exactly how I did it. Well-written solution. Thank you for posting it.

Hosam Hajjir - 1 year, 7 months ago

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Thanks! This was a great problem!

David Vreken - 1 year, 7 months ago

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