Regular Octagon

The essence of mathematics is its freedom. — Georg Cantor

## 9L solution

Let point A and point B be the two given points.

1. Construct the perpendicular to line AB through point A. Let point C be an arbitrary point on this new line (C below AB to match orientation of explore mode).
2. Construct the angle bisector of BAC. Let point D be an arbitrary point on this line.
3. Construct the perpendicular to line AD through point B. Let point E be an arbitrary point on this line on the opposite side of point B from point C.
4. Construct the angle bisector of ABE. This line intersects with line AD at point F and line AC at point G.
5. Construct the perpendicular to line AB through point F.
6. Construct the perpendicular to line AG through point G. This line intersects with line BE at point H.
7. Construct the perpendicular bisector of point B and point H. This line intersects with line BG at point I.
8. Construct the perpendicular of line AB through point I.
9. Construct the perpendicular of line AD through point G (total 29E).

## 13E solution

Let A and B be the two given points.

1. Construct circle A with center A and radius AB.
2. Construct circle B with radius AB. Circle B intersects with circle A at points C and D.
3. Construct line CD. Line CD intersects with line AB at point E.
4. Construct circle E with center E and radius AE. Circle E intersects with CD at two points; let one of these two points be point F (F below AB to match the orientation of explore mode).
5. Construct line AF. Line AF intersects with circle A at two points; let the point opposite of point A from point F be point G.
6. Construct line BF. Line BF intersects with circle B at two points; let the point opposite of point B from point F be point H.
7. Construct circle F with radius FG.
	* Circle F intersects with line AF at two points; let the point opposite of F from A be point I.
	* Circle F intersects with line BF at two points; let the point opposite of F from B be point J.
	* Circle F intersects with line EF at two points; let the point opposite of E from F be point K.
8. Construct line HI.
9. Construct line GJ.
10. Construct circle K with center K and radius FK.
	* Circle K intersects with line FG at two points; one is point F, the other is point L.
	* Circle K intersects with line GJ at two points; let the point opposite of G from J be point M.
	* Circle K intersects with line FK at two points; one is point F, the other is point N.
	* Circle K intersects with line HI at two points; let the point opposite of H from I be point O.
	* Circle K intersects with line FH at two points; one is point F, the other is point P.
11. Construct line LN.
12. Construct line NP.
13. Construct line MO (total 13L).

## 2V solution

Mirror any solution over line AB. Note that two of the sides were already constructed, leaving only 5 remaining sides to go. The following constructions minimize moves:

From the 9L solution, with point J as the intersection of the perpendicular of AB through F with the bisector of EH, with point K as the intersection of the perpendicular of AB through I with the perpendicular of AD through G, and with point L as the intersection of BE with IK:

1. Construct line AJ.
2. Construct line BK.
3. Construct the perpendicular to line AJ through point L, intersecting line AC at M.
4. Construct the perpendicular to line AM through point M.
5. Construct the perpendicular to line AF through point F (total 14L 40E).

Mirroring the steps for sides missing from the 13E solution results in 20L 20E (at least two more auxiliary lines or circles are required). But it is also possible to interleave the construction of both octagons for a 2V solution with even fewer moves:

1. Construct circle A with center A and radius AB.
2. Construct circle B with radius AB. Circle B intersects with circle A at points C and D.
3. Construct line CD. Line CD intersects with line AB at point E.
4. Construct circle E1 with center E and radius AE. Circle E1 intersects with CD at F and G.
5. Construct line AG, with point H at the intersection of circle A furthest from G.
6. Construct line AF, with point I at the intersection of circle A furthest from F.
7. Construct line BF, with point J at the intersection of circle B furthest from F.
8. Construct line BG, with point K at the intersection of circle B furthest from G.
9. Construct line HI.
10. Construct line JK, with point L at the intersection of AG.
11. Construct circle E2 with center E and radius EL, creating points M through S as follows:
	* Circle E2 intersects line AG at L and Q;
	* Circle E2 intersects line BF at M and P, M closest to L;
	* Circle E2 intersects line BG at N and S, N closest to M;
	* Circle E2 intersects line AF at O and R, O closest to N.
12. Construct line LM.
13. Construct line NO.
14. Construct line PQ.
15. Construct line RS.
16. Construct line BL, intersecting PQ at T.
17. Construct circle E3 with center E and radius ET, creating points U, V and W as follows:
	* Circle E3 intersects line NO at two points, U closest to T;
	* Circle E3 intersects line LM at two points, V closest to L;
	* Circle E3 intersects line RS at two points, W closest to S.
18. Construct line TU.
19. Construct line VW (total 19L 19E).

## Explanation

For the 9L solution, the emphasis is on producing the remaining 7 sides with as few extraneous moves as possible. In a regular octagon, all internal angles are 135°, and external angles are 45°. Like all regular polygons with an even number of sides, any interior angle bisector or perpendicular from a vertex intersects with another vertex. Thus, the first two moves produce a 45° angle as fast as possible, and while the first four moves provide only two sides, they also serve to locate point G as one of the other vertices, as well as external point F and later I that can be used as the apex of an isosceles triangle to get other sides. Every remaining move then adds another side, favoring tools with 3E cost (several of the moves could be replaced by a use of the Parallel tool, but at a cost of 4E).

For the 13E solution, obtaining the first 45° angle is done with the midpoint E located by the perpendicular bisector of AB, although it must be done with elementary moves rather than the bisector tool because circles A and B also identify additional vertices G and H. Circle F then circumscribes square GHIJ from its diagonals to get two more sides, as well as identifying point K as the center of the octagon. Finally, circle K (congruent to F) circumscribes two squares each coincident with alternating sides of the octagon, and all that remains is to trace those sides.