## Classical cube problem

(FB/Math/BM)

Let two cubes C₁ and C₂ have edge lengths a and b respectively. A hole is drilled, carved or whatever through C₁ so that C₂ can pass through. No restrictions on the form of the hole except that it should really be a hole (topologically speaking the remainder should have genus 1).

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The cube with edge a is projected onto a plane along a space diagonal (length a√3). The projection is a regular hexagon and the lengths of the projections of the edges are a√(2/3). Then imagine a hole cut along …the same direction (normal to the plane). In the figure this shows up as a square with side b. Now, let φ = 15⁰, so that tan φ = 2 – √3. In this position the square has all its corners on a side of the hexagon.

The upper right side of the hexagon has equation
y = -√3*(x – a√(2/3)) and the the line from the origin at angle φ has equation y = (2 – √3)x. From this the corner of the square has coordinates
a/√2·(1,2 – √3) and so its length is a√( 4 – 2√3). The upper corner has the same distance to the origin so the position inside the hexagon is possible. Finally
b = a√( 4 – 2√3) * √2 = 2a√( 2 – √3) ≈ 1.035a.

So a cube can pass a hole through a somewhat *smaller* cube!

Annonser