Regular Hexagon

If only I had the theorems! Then I should find the proofs easily enough. — Bernard Riemann

## Level Information

### Instruction:

Construct a regular hexagon with the given side.

## 7L solution

Given the side AB.

1. Construct the circle with center A and radius AB
2. Construct the circle with center B and radius AB, intersecting circle A at C and D (C below AB to match the orientation of explore mode)
3. Construct line AC, intersecting circle A at E
4. Construct line BC, intersecting circle B at F
5. Construct the parallel to AC through F
6. Construct the parallel to BC through E, intersecting the the parallel to AC at G
7. Construct the perpendicular bisector of DG (total 15E)

## 8E solution

Given the side AB.

1. Construct the circle with center A and radius AB
2. Construct the circle with center B and radius AB, intersecting circle A at C and D (C below AB to match the orientation of explore mode)
3. Construct the circle with center D and radius CD, intersecting circle A at E and circle B at F
4. Construct line AC, intersecting circle A at G and circle D at H
5. Construct line BC, intersecting circle B at I and circle D at J
6. Construct line EG
7. Construct line FI
8. Construct line HJ

## 2V solution

By reusing artifacts from the first hexagon, only 5 additional lines are needed for a 2V solution, regardless of whether you start from the 7L or 8E solution. From the 7L solution, with H as the intersection of line EG and circle A, and I as the intersection of line FG and circle B:

1. Construct line AD, intersecting circle A at J
2. Construct line BD, intersecting circle B at K
3. Construct line JH, intersecting line BF at L
4. Construct line IK, intersecting line AE at M
5. Construct line LM (total 12L 20E; would be 13L 13E with similar constructions from the 8E solution)

It is also possible to construct a 2V solution in 13E with only 2 circles, by constructing line AB in place of circle D in the 8E solution, although it requires interleaving of steps between the two hexagons.

## Explanation

The internal angle at each vertex of a regular hexagon is 120°. Both solutions start by constructing equilateral triangle ABC to establish a 60° external angle below AB, which in turn gives the desired 120° angle for isosceles triangles BAE and ABF forming the second and third sides of the hexagon. From there, the 7L solution minimizes moves by directly constructing the remaining sides: the fourth and fifth are easy because a vertex is already present and opposite sides are parallel (the angle bisector tool can be used instead of the parallel tool, but with no difference in E-cost). The final side has no starting vertex, but it can be observed that EFG is an equilateral triangle with sides twice as long as AB, with D at the midpoint of its base, and therefore the perpendicular bisector of DG gives the remaining side.

Meanwhile, the 8E solution notes that point D is the center of the hexagon to be constructed. For each of the 6 equilateral triangles that make up one side of the interior of the target hexagon (such as ABD), a corresponding equilateral triangle (such as ABC) completes a rhombus that has its final vertex on circle D (such as ACBD). This gives the location of both points E and F for the fourth and fifth sides, as well as the inscribed equilateral triangle CHJ sufficient to trace the final side.

Any 2V solution reflects the first hexagon across line AB.