# How can I prove the existence of an octagon/decagon/dodecagon?

I've had this question in my head recently due to my math teacher giving me this problem as a bonus on a test.

So I have two regular hexagons inscribed in two distinct circles, of radius n. The two circles intersect, the intersection points being vertices of the hexagons. Let them be O1 and O2.

Part (a) of the question was to prove the existence of a circle of radius n that passes through 2 vertices of the first hexagon and 2 vertices of the second, while containing only ONE of the intersection points.

So suppose that this circle intersects one hexagon at vertices A and B and the other at vertices C and D. This circle, while intersecting at A, B, C, and D, would also have to include one of O1, O2, in its interior.

Part (b) of the question was to prove/disprove that these 4 points (A, B, C, D) can make up the vertices of a regular octagon/decagon/dodecagon.

Nobody in my class, including myself, got the bonus.

My teacher also proposed the same problem, but with pentagons and heptagons, and to prove/disprove existence of hexagon/octagon/decagon for pentagon, and decagon/dodecahedron/14-gon for the heptagon.

• Could you elaborate? – user181475 Oct 22 '14 at 9:51
• I understand that Roots of Unity can be used to define a regular polygon. I'm still not seeing how we can use Roots of Unity in this problem, however. – user181475 Oct 22 '14 at 9:54
• dodecahedron or dodecagon? – Hagen von Eitzen Oct 22 '14 at 10:26
• Dodecagon, fixed. – user181475 Oct 23 '14 at 12:19
• This seems like a specific case of this recent question. As of the writing of this comment, the question is locked, and the answers "soft-deleted", because the problem is from an on-going contest. Once the contest ends, you should find some insights there. – Blue Oct 31 '14 at 5:54

For the first part - draw a diagram. Note that the side of a hexagon inscribed in a circle of radius $r$ is itself $r$ (equilateral triangles). Take one of the common points of the two hexagons as point $A$ and the other as point $D$. The other two points adjacent to $A$ are labelled $B$ and $C$ - one on each hexagon. Note that the angle $BAC$ is equal to $120^{\circ}$.
There is a circle which passes through the three points $ABC$ (this is true for any three points in a plane which do not lie on a straight line). Call the centre of this circle $O$ so that $OA=OB=OC$. By symmetry angle $OAB=60^{\circ}$. An isosceles triangle with one angle equal to $60^{\circ}$ is equilateral.
For the second part $A$ is in the interior of triangle $BCD$ so the four points cannot lie on the boundary of any convex figure.