Equilateral Triangle in Circle

We do not choose mathematics as our profession; it chooses us. — Yuri Manin

## Level Information

### Instruction:

Inscribe an equilateral triangle in the circle using the given point as a vertex. The center of the circle is not given.

## 5L solution

Let A be the vertex.

1. Construct a circle with center A and an arbitrary radius, intersecting the original circle at B and C
2. Construct the circle with center B and radius AB, intersecting circle A at D and E
3. Construct the perpendicular bisector of CD, intersecting the original circle at F
4. Construct the perpendicular bisector of CE, intersecting the original circle at G
5. Construct line FG (total 9E)

Alternate solution:

1. Construct a circle with center A and an arbitrary radius, intersecting the original circle at B and C
2. Construct the perpendicular bisector of AC, intersecting the second circle at D and E
3. Construct the perpendicular bisector of BD, intersecting the original circle at F
4. Construct the perpendicular bisector of BE, intersecting the original circle at G
5. Construct line FG (total 11E)

## 6E solution

Let A be the vertex.

1. Construct a circle with an arbitrary center B on the circle and radius AB, intersecting the circle at C
2. Construct the circle with center A and radius AC, intersecting circle B at D
3. Construct the circle with center D and radius AD, intersecting circle A at E and F
4. Construct line AE, intersecting the circle at G
5. Construct line AF, intersecting the circle at H
6. Construct line GH

## Explanation

In both 5L solutions, triangle ABC is isosceles by construction, with points B and C reflections of one another across the line from A through the center of the circle, even if that line is not drawn. The next step constructs enough information to also go through the center of the circle (the first solution by drawing a second circle where connecting points D and E would be the perpendicular bisector of a chord AB, the second solution by directly bisecting the chord AC to determine where points D and E lie), while also determining three points that form a pair of equilateral triangles (ABD and ABE for the first solution, ACD and ACE for the second solution). The next two steps take the perpendicular bisector of the reflected point on circle A through points D and E; the angle bisector of the two points with A as the vertex would produce the same effect, but with a higher E cost. Regardless of the arc angle of circle A subtended by CDB, constructing the angle bisector after subtracting out the 60° arc along the equilateral triangle means that the resulting bisector will be offset by half of that, or 30°, from the line of symmetry in isosceles triangle ABC, which provides the needed angle for two of the sides of the final triangle. The third side FG is then trivial to construct.

The 6E solution starts by determining a point D that lies along the tangent to the circle at point A; this can be seen by realizing triangle ACD is isosceles, symmetric across line AB, and that arc AB and BC along the original circle are also identical in arc length. The next step constructs circle D, such that triangles ADE and ADF are both equilateral, giving the two sides of the desired triangle through point A. Inscribing the third side GH is then trivial.