Revisions to Surface dissection of regular tetrahedron to cube

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edited Apr 13, 2017 at 12:58

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Does anyone know what is the fewest-piece dissection of the surface of a regular tetrahedron to the surface of a cube (of the same area)?

It is well-known that the volume of a regular tetrahedron cannot be dissected to the volume of a cube, because their Dehn invariants differ. But their surfaces have a dissection, by applying the Boyai-Gerwien theorem. Applying that theorem in one way (among several options) leads to an $31$-piece dissection (if I calculated correctly), illustrated below. But this is surely very far from optimal (and hardly aesthetically pleasing). Likely the question has been explored, but I have not found any literature. It could make a pleasing contrast to the impossibility of a volume dissection.

Let the cube edge length be $1$, so its surface area is $6$. A tetrahedron edge length of $L= 2^{\frac{1}{2}} 3^{\frac{1}{4}} \approx 1.86$ leads to a surface area of $\sqrt{3} L^2 = 6$.

^{Surface dissection of regular tetrahedron to cube.}

Does anyone know what is the fewest-piece dissection of the surface of a regular tetrahedron to the surface of a cube (of the same area)?

It is well-known that the volume of a regular tetrahedron cannot be dissected to the volume of a cube, because their Dehn invariants differ. But their surfaces have a dissection, by applying the Boyai-Gerwien theorem. Applying that theorem in one way (among several options) leads to an $31$-piece dissection (if I calculated correctly), illustrated below. But this is surely very far from optimal (and hardly aesthetically pleasing). Likely the question has been explored, but I have not found any literature. It could make a pleasing contrast to the impossibility of a volume dissection.

Let the cube edge length be $1$, so its surface area is $6$. A tetrahedron edge length of $L= 2^{\frac{1}{2}} 3^{\frac{1}{4}} \approx 1.86$ leads to a surface area of $\sqrt{3} L^2 = 6$.

^{Surface dissection of regular tetrahedron to cube.}

Does anyone know what is the fewest-piece dissection of the surface of a regular tetrahedron to the surface of a cube (of the same area)?

It is well-known that the volume of a regular tetrahedron cannot be dissected to the volume of a cube, because their Dehn invariants differ. But their surfaces have a dissection, by applying the Boyai-Gerwien theorem. Applying that theorem in one way (among several options) leads to an $31$-piece dissection (if I calculated correctly), illustrated below. But this is surely very far from optimal (and hardly aesthetically pleasing). Likely the question has been explored, but I have not found any literature. It could make a pleasing contrast to the impossibility of a volume dissection.

Let the cube edge length be $1$, so its surface area is $6$. A tetrahedron edge length of $L= 2^{\frac{1}{2}} 3^{\frac{1}{4}} \approx 1.86$ leads to a surface area of $\sqrt{3} L^2 = 6$.

^{Surface dissection of regular tetrahedron to cube.}

Typo. 1.81 should have been 1.86.

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edited Feb 27, 2017 at 20:11

Joseph O'Rourke

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Does anyone know what is the fewest-piece dissection of the surface of a regular tetrahedron to the surface of a cube (of the same area)?

It is well-known that the volume of a regular tetrahedron cannot be dissected to the volume of a cube, because their Dehn invariants differ. But their surfaces have a dissection, by applying the Boyai-Gerwien theorem. Applying that theorem in one way (among several options) leads to an $31$-piece dissection (if I calculated correctly), illustrated below. But this is surely very far from optimal (and hardly aesthetically pleasing). Likely the question has been explored, but I have not found any literature. It could make a pleasing contrast to the impossibility of a volume dissection.

Let the cube edge length be $1$, so its surface area is $6$. A tetrahedron edge length of $L= 2^{\frac{1}{2}} 3^{\frac{1}{4}} \approx 1.81$$L= 2^{\frac{1}{2}} 3^{\frac{1}{4}} \approx 1.86$ leads to a surface area of $6$$\sqrt{3} L^2 = 6$.

^{Surface dissection of regular tetrahedron to cube.}

Does anyone know what is the fewest-piece dissection of the surface of a regular tetrahedron to the surface of a cube (of the same area)?

It is well-known that the volume of a regular tetrahedron cannot be dissected to the volume of a cube, because their Dehn invariants differ. But their surfaces have a dissection, by applying the Boyai-Gerwien theorem. Applying that theorem in one way (among several options) leads to an $31$-piece dissection (if I calculated correctly), illustrated below. But this is surely very far from optimal (and hardly aesthetically pleasing). Likely the question has been explored, but I have not found any literature. It could make a pleasing contrast to the impossibility of a volume dissection.

Let the cube edge length be $1$, so its surface area is $6$. A tetrahedron edge length of $L= 2^{\frac{1}{2}} 3^{\frac{1}{4}} \approx 1.81$ leads to a surface area of $6$.

^{Surface dissection of regular tetrahedron to cube.}

Does anyone know what is the fewest-piece dissection of the surface of a regular tetrahedron to the surface of a cube (of the same area)?

It is well-known that the volume of a regular tetrahedron cannot be dissected to the volume of a cube, because their Dehn invariants differ. But their surfaces have a dissection, by applying the Boyai-Gerwien theorem. Applying that theorem in one way (among several options) leads to an $31$-piece dissection (if I calculated correctly), illustrated below. But this is surely very far from optimal (and hardly aesthetically pleasing). Likely the question has been explored, but I have not found any literature. It could make a pleasing contrast to the impossibility of a volume dissection.

Let the cube edge length be $1$, so its surface area is $6$. A tetrahedron edge length of $L= 2^{\frac{1}{2}} 3^{\frac{1}{4}} \approx 1.86$ leads to a surface area of $\sqrt{3} L^2 = 6$.

^{Surface dissection of regular tetrahedron to cube.}

Cleaned up some minor flaws in the figure.

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edited Feb 24, 2017 at 21:02

Joseph O'Rourke

150.8k
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358
958

Does anyone know what is the fewest-piece dissection of the surface of a regular tetrahedron to the surface of a cube (of the same area)?

It is well-known that the volume of a regular tetrahedron cannot be dissected to the volume of a cube, because their Dehn invariants differ. But their surfaces have a dissection, by applying the Boyai-Gerwien theorem. Applying that theorem in one way (among several options) leads to an $31$-piece dissection (if I calculated correctly), illustrated below. But this is surely very far from optimal (and hardly aesthetically pleasing). Likely the question has been explored, but I have not found any literature. It could make a pleasing contrast to the impossibility of a volume dissection.

Let the cube edge length be $1$, so its surface area is $6$. A tetrahedron edge length of $L= 2^{\frac{1}{2}} 3^{\frac{1}{4}} \approx 1.81$ leads to a surface area of $6$.

^{Surface dissection of regular tetrahedron to cube.}

Does anyone know what is the fewest-piece dissection of the surface of a regular tetrahedron to the surface of a cube (of the same area)?

It is well-known that the volume of a regular tetrahedron cannot be dissected to the volume of a cube, because their Dehn invariants differ. But their surfaces have a dissection, by applying the Boyai-Gerwien theorem. Applying that theorem in one way (among several options) leads to an $31$-piece dissection (if I calculated correctly), illustrated below. But this is surely very far from optimal (and hardly aesthetically pleasing). Likely the question has been explored, but I have not found any literature. It could make a pleasing contrast to the impossibility of a volume dissection.

Let the cube edge length be $1$, so its surface area is $6$. A tetrahedron edge length of $L= 2^{\frac{1}{2}} 3^{\frac{1}{4}} \approx 1.81$ leads to a surface area of $6$.

^{Surface dissection of regular tetrahedron to cube.}

Does anyone know what is the fewest-piece dissection of the surface of a regular tetrahedron to the surface of a cube (of the same area)?

It is well-known that the volume of a regular tetrahedron cannot be dissected to the volume of a cube, because their Dehn invariants differ. But their surfaces have a dissection, by applying the Boyai-Gerwien theorem. Applying that theorem in one way (among several options) leads to an $31$-piece dissection (if I calculated correctly), illustrated below. But this is surely very far from optimal (and hardly aesthetically pleasing). Likely the question has been explored, but I have not found any literature. It could make a pleasing contrast to the impossibility of a volume dissection.

Let the cube edge length be $1$, so its surface area is $6$. A tetrahedron edge length of $L= 2^{\frac{1}{2}} 3^{\frac{1}{4}} \approx 1.81$ leads to a surface area of $6$.

^{Surface dissection of regular tetrahedron to cube.}

added 36 characters in body

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edited Feb 24, 2017 at 16:07

Joseph O'Rourke

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asked Feb 24, 2017 at 15:57

Joseph O'Rourke

150.8k
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