Solution to "Cutting A Hole In Half" from 12 Apr 26
Monday 13 April 2026 at 21:36
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The problem concerned a shape consisting of a rectangle with a rectangular hole in it and asked fo with one straight cut we can divide the shape into two parts whose areas are the same.
The solution is to cut along the line that passes through the centre of the rectangle and the centre of the hole.
This works because a rectangle has central symmetry; a rotation of 180° about its centre maps it to itself; and therefore a line through the centre cuts it into two parts which must be congruent, because a rotation of 180° about its centre simply exchanges the two parts with one another.
In our problem the cut goes through both centres and must therefore divide the rectangle into congruent parts and the hole into congruent parts, and so the two parts the shape is cut into each consist of precisely half the rectangle minus precisely half the hole. Consequently their areas are equal, as required.
A similar idea ought to work for any centrally symmetric figure with centrally symmetric hole, such as a decagon and a circle, or hexagon and a rhombus etc. etc.
Even more interesting and perhaps rather surprising, is that given any three 3D solids that have central symmetry—say a sphere , a cube and a dodecahedron—in arbitrary position, we can simultaneously divide all three into identical halves by a single plane cut. We just choose the plane that passes through all three centres.
Solution to "Cutting A Hole In Half" from 12 Apr 26
The problem concerned a shape consisting of a rectangle with a rectangular hole in it and asked fo with one straight cut we can divide the shape into two parts whose areas are the same.
The solution is to cut along the line that passes through the centre of the rectangle and the centre of the hole.
This works because a rectangle has central symmetry; a rotation of 180° about its centre maps it to itself; and therefore a line through the centre cuts it into two parts which must be congruent, because a rotation of 180° about its centre simply exchanges the two parts with one another.
In our problem the cut goes through both centres and must therefore divide the rectangle into congruent parts and the hole into congruent parts, and so the two parts the shape is cut into each consist of precisely half the rectangle minus precisely half the hole. Consequently their areas are equal, as required.
A similar idea ought to work for any centrally symmetric figure with centrally symmetric hole, such as a decagon and a circle, or hexagon and a rhombus etc. etc.
Even more interesting and perhaps rather surprising, is that given any three 3D solids that have central symmetry—say a sphere , a cube and a dodecahedron—in arbitrary position, we can simultaneously divide all three into identical halves by a single plane cut. We just choose the plane that passes through all three centres.