Euclid’s Elements — Book XIII, Proposition 14

Constructing a Regular Octahedron Inside a Sphere

In Book XIII of Euclid’s Elements, Proposition 14 explains how to construct a regular octahedron and inscribe it perfectly within a given sphere.

It also establishes a beautiful relationship between the diameter of the sphere and the side of the octahedron.

This proposition is one of the key steps in Euclid’s exploration of the Platonic solids.

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Proposition Statement

To construct a regular octahedron, inscribe it in a given sphere, and prove that the square on the sphere’s diameter is double the square on the side of the octahedron.

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Geometric Setup

1. Start with a sphere having a given diameter AB.

2. Bisect AB at C.

3. Construct a semicircle ADB on AB and draw CD perpendicular to AB.

4. Set up a square EFGH with each side equal to DB.

5. From the square’s center K, erect a perpendicular KM to the plane of the square.

6. Choose KL = KE, where KE is half the diagonal of the square.

7. Connect the vertices L and M to all four vertices of the square E, F, G, H.

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Key Results

1. Construction of the Regular Octahedron

• By construction, the triangles LEH, LHF, LFG, LGE, and their counterparts at M are all equilateral.

• Thus, the resulting solid is a regular octahedron:

  • 8 faces → equilateral triangles,

  • 12 edges → all equal,

  • 6 vertices → perfectly symmetric.

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2. The Octahedron Is Inscribed in the Sphere

• Using symmetry, Euclid proves that all vertices of the octahedron (E, F, G, H, L, M) lie on the sphere’s surface.

• Therefore, the octahedron is perfectly inscribed.

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3. Relationship Between the Diameter and the Octahedron’s Side

Euclid demonstrates that:

$$A{B}^2=2\cdot L{E}^2$$

where:

• AB = diameter of the sphere,

• LE = side of the octahedron.

Equivalently:

$$LE=\frac{AB}{\sqrt{\mathrm{2}}}$$

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Sketch of the Proof

1. Since AB is bisected at C, we have $AB = 2 \cdot BC$.
   Thus, $\dfrac{AB}{BC} = 2$, leading to:

   $$A{B}^2=2\cdot B{D}^2\mathrm{.}$$

2. From the construction of the square and perpendicular, DB = EH = LE, so:

   $$A{B}^2=2\cdot L{E}^2\mathrm{.}$$

3. Therefore, the square on the diameter is double the square on the octahedron’s side.

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Mathematical Significance

This proposition provides:

• A geometric construction of the regular octahedron.

• A precise formula relating the sphere’s diameter and the edge length of the octahedron.

• A foundation for later propositions involving the icosahedron and dodecahedron.

In modern notation, if the circumsphere radius is $R$, the side $a$ of a regular octahedron satisfies:

$$R=\frac{a}{\sqrt{2}}\text{or equivalently}a=R\sqrt{2}\mathrm{.}$$

This is consistent with Euclid’s result.

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Summary

Euclid constructs a regular octahedron inscribed in a sphere and proves that:

$$\boxed{{\displaystyle A{B}^2=2\cdot L{E}^{\mathrm{2}}}}$$

where:

• AB = diameter of the sphere,

• LE = side of the octahedron.

Thus, the octahedron fits perfectly inside the sphere, and its edge length relates beautifully to the sphere’s size.

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Why This Matters

The regular octahedron is one of the five Platonic solids and plays a key role in Euclid’s systematic study of symmetrical 3D figures.

This proposition elegantly links 3D geometry with circle and square constructions, showcasing Euclid’s mastery in uniting planar and spatial reasoning.