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 ($\text {940}$ – $\text {998}$): $38$