Inscribing Equilateral Triangle inside Square with a Coincident Vertex/Mistake

Source Work

 * The Puzzles:
 * Abul Wafa : $38$
 * Solution
 * Solution

Mistake

 * Join $B$ to the midpoint, $M$, of $DC$. Draw an arc with centre $B$ and radius $BA$ to cut $BM$ at $N$. Let $DN$ cut $CB$ at $H$. Then $H$ is one of the vertices sought.


 * Inscribing-equilateral-triangle-inside-square-mistake.png

Correction
The construction is incorrect.

$DH$ is shorter than $GH$.

Let $\Box ABCD$ be embedded in a Cartesian plane such that:

By Equation of Straight Line in Plane and some algebra, the equation for the straight line $MB$ is:
 * $(1): \quad y = 2 \paren {a - x}$

By Equation of Circle in Cartesian Plane and some algebra, the equation for the circle with center at $B$ and radius $a$ is:
 * $(2): \quad y^2 = 2 a x - x^2$

Hence their point of intersection $M$ is found by solving the simultaneous equations $(1)$ and $(2)$:

We are interested in the negative square root in this expression, as the positive square root corresponds to the point on $BM$ for negative $y$.

Thus we have:

Thus we can calculate the tangent of $\angle CDH$:

But from Tangent of 15 Degrees:


 * $\tan 15 \degrees = 2 - \sqrt 3$

So:
 * $\angle CDH \ne \tan 15 \degrees$

(in fact it is about $10.8 \degrees$).

Hence from Inscribing Equilateral Triangle inside Square with a Coincident Vertex: Lemma:
 * $\triangle DGH$ is not an equilateral triangle.

It is not clear where the error originates, whether:
 * in the original of
 * in 's work
 * in 's work.