Euclid’s Elements — Book XIII, Proposition 13

Constructing a Tetrahedron Inside a Sphere

In Book XIII of Euclid’s Elements, Proposition 13 marks an important step in the study of the Platonic solids.

Here, Euclid constructs a regular tetrahedron (a triangular pyramid with four equilateral faces) so that it is perfectly inscribed in a given sphere.

He also establishes a beautiful relationship between the sphere’s diameter and the side of the tetrahedron.

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Proposition Statement

To construct a regular tetrahedron inscribed in a given sphere, and to prove that the square on the diameter of the sphere is one and a half times the square on the side of the tetrahedron.

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Geometric Setup

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

2. On AB, choose point C so that:

   $$AC=2\cdot CB\mathrm{.}$$

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

4. On a separate plane, draw a circle with radius equal to CD and inscribe an equilateral triangle EFG.

5. Erect a perpendicular HK at the center of the triangle’s plane and set:

   $$HK=AC\mathrm{.}$$

6. Join KE, KF, KG to form a pyramid with EFG as the base and K as the apex.

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Key Results

Euclid proves two things:

1. The Tetrahedron Is Perfectly Inscribed in the Sphere

• The four equilateral triangles EFG, KEF, KFG, and KEG form a regular tetrahedron.

• Using properties of semicircles and right angles, Euclid shows that all vertices E, F, G, K lie on the surface of the sphere.

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2. Relationship Between the Diameter and the Pyramid’s Side

Euclid establishes the following identity:

$$A{B}^2=\frac{3}{2}\cdot A{D}^2,$$

where:

• AB = diameter of the sphere,

• AD = side of the tetrahedron.

Thus, the square on the sphere’s diameter is 1.5 times the square on the side of the tetrahedron.

Equivalently:

$$AD=\frac{AB}{\sqrt{\frac{3}{2}}}=AB\cdot \sqrt{\frac{2}{3}}\mathrm{.}$$

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Sketch of the Proof

Euclid’s reasoning follows these key steps:

1. Relating Diameter Segments

   • Since $AC = 2 \cdot CB$, we have:

     $$AB=3\cdot CB\mathrm{.}$$

2. Using a Mean Proportional

   • By a lemma, Euclid proves:

     $$\frac{AB}{BC}=\frac{A{D}^2}{D{C}^2},$$

     establishing proportionality between the sphere’s diameter and the tetrahedron’s dimensions.

     

3. Equilateral Triangle Property

   • From Proposition XIII.12, the side of the equilateral triangle EF relates to the circle’s radius through:

     $$E{F}^2=3\cdot E{H}^2\mathrm{.}$$

   • Since DC = EH, this leads to the equivalence between DA and EF.

4. Concluding the Ratio

   • Substituting the relationships, Euclid concludes:

     $$A{B}^2=\frac{3}{2}\cdot A{D}^2\mathrm{.}$$

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Mathematical Significance

This proposition is fundamental because it provides:

• A construction for a regular tetrahedron inscribed in a sphere.

• An exact formula linking the side length of the tetrahedron to the sphere’s diameter.

• A stepping stone to later results in Book XIII, which explore the geometry of the icosahedron and dodecahedron.

In modern terms, this directly connects the circumsphere radius $R$ of a regular tetrahedron to its edge length $a$:

$$R=\frac{a\sqrt{6}}{4},$$

which is equivalent to Euclid’s result.

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Summary

Euclid constructs a regular tetrahedron inside a given sphere and proves that:

$$\boxed{{\displaystyle A{B}^2=\frac{3}{2}\cdot A{D}^{\mathrm{2}}}}$$

where:

• AB = diameter of the sphere,

• AD = side of the tetrahedron.

Thus, the tetrahedron is perfectly inscribed, and its edge length relates beautifully to the sphere’s size.