Logic is the anatomy of thought. — John Locke
5L solution
Let O be the center of the circle and A the point on it.
- Construct line OA, intersecting the circle at B
- Construct the perpendicular to OA from A
- Construct the circle with center B and radius OB, intersecting the circle at C and D, and line OA at E
- Construct line CE
- Construct line DE
6E solution
Let O be the center of the circle and A the point on it.
- Construct line OA, intersecting the circle at B
- Construct the circle with center B and radius OB, intersecting the circle at C and D, and line OA at E
- Construct the circle with center O and radius OE
- Construct line CE, intersecting the latest circle at F
- Construct line DE, intersecting the latest circle at G
- Construct line FG
Explanation
Line OA forms an altitude of the equilateral triangle, with A at the midpoint of the base. Both constructions use circle B to determine the two sides at a 30° angle from the apex; this can be seen since triangles BCO and BDO are equilateral by construction, and angles ECO and EDO are 90 by Thales' theorem. Having drawn tangents through C and D, point E is the apex of the desired triangle. The final leg of the triangle through A is built with the perpendicular tool for the 5L solution, and by inscribing an equilateral triangle into the second larger circle O in the 6E solution.