Inscribing Equilateral Triangle inside Square with a Coincident Vertex

Theorem

Let $\Box ABCD$ be a square.

It is required that $\triangle DGH$ be an equilateral triangle inscribed within $\Box ABCD$ such that vertex $D$ of $\triangle DGH$ coincides with vertex $D$ of $\Box ABCD$.

Construction 1

By Construction of Equilateral Triangle, let an equilateral triangle $\triangle ABN$ be constructed on $AB$ such that $N$ is inside $\Box ABCD$.

Let $AB$ be produced to $F$ such that $AB = BF$.

Draw an arc centred at $F$ with radius $FN$ to cut $AB$ at $G$.

Construct $H$ on $BC$ such that $DH = DG$.

Then $DGH$ is the required equilateral triangle.

Construction 2

By Construction of Equilateral Triangle, let an equilateral triangle $\triangle DAI$ be constructed on $AD$ such that $I$ is inside $\Box ABCD$.

Bisect $\angle ADI$ and bisect it again towards $AD$ to cut $AB$ at $G$.

Construct $H$ on $BC$ such that $DH = DG$.

Then $DGH$ is the required equilateral triangle.

Construction 3

By Construction of Equilateral Triangle, let an equilateral triangle $\triangle CDN$ be constructed on $CD$ such that $N$ is inside $\Box ABCD$.

Bisect $\angle ADN$ to cut $AB$ at $G$.

Construct $H$ on $BC$ such that $DH = DG$.

Then $DGH$ is the required equilateral triangle.

Construction 4

By Construction of Equilateral Triangle, let an equilateral triangle $\triangle ABN$ be constructed on $AB$ such that $N$ is inside $\Box ABCD$.

Let $DN$ be produced to cut $BC$ at $H$.

Construct $G$ on $AB$ such that $DH = DG$.

Then $DGH$ is the required equilateral triangle.

Historical Note

This problem was raised by Abu'l-Wafa Al-Buzjani, who gave $5$ different constructions.

Sources

• 1992: David Wells: Curious and Interesting Puzzles ... (previous) ... (next): Abul Wafa ($\text {940}$ – $\text {998}$): $38$