The Pythagorean Hexagon — Proof and Definition

The Pythagorean Hexagon — Proof and Definition
© Tiago Hands — 18 October 2025 (UTC +8)

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Definition and Construction

Six points A, B, C, D, E and F form a hexagon whose sides are all equal in length.
Point G lies at the centre of a circle with radius AB = x > 0.
Points F, B and D lie on the circumference of this circle, forming a right-angled triangle at G where angle FGB = 90°.

Quadrilateral ABGF forms a square.
Quadrilaterals FGDE and BCDG form rhombuses that share the same side length x.
The internal angles satisfy α + β = 270°.

Each figure has a centre:
H for square ABGF, I for rhombus BCDG, and J for rhombus FGDE.
When these three figures are joined edge-to-edge around the circle, their areas obey a Pythagorean relation.

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Areas of the Component Figures

Let AB = x.

Square ABGF: A₁ = x²
Rhombus FGDE: A₂ = 2x²cos(β/2)sin(β/2) = x²sin(β)
Rhombus BCDG: A₃ = 2x²cos(α/2)sin(α/2) = x²sin(α)

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Algebraic and Trigonometric Proof

We aim to show A₂² + A₃² = A₁².

Substituting:
(x²sin(β))² + (x²sin(α))² = x⁴(sin²(β) + sin²(α))

Since α + β = 270°, α = 270° − β, and sin(270° − β) = −cos(β),
we obtain x⁴(sin²(β) + cos²(β)) = x⁴.

Hence, A₂² + A₃² = A₁².

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Geometric Implication

The areas satisfy the Pythagorean relation:
A₁² = A₂² + A₃²,
where A₁ is the area of square ABGF, A₂ is the area of rhombus FGDE, and A₃ is the area of rhombus BCDG.

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End of Proof