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.

I've thought about this question for a while, and I've given up.

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.

  • $\begingroup$ Could you elaborate? $\endgroup$ – user181475 Oct 22 '14 at 9:51
  • $\begingroup$ 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. $\endgroup$ – user181475 Oct 22 '14 at 9:54
  • $\begingroup$ dodecahedron or dodecagon? $\endgroup$ – Hagen von Eitzen Oct 22 '14 at 10:26
  • $\begingroup$ Dodecagon, fixed. $\endgroup$ – user181475 Oct 23 '14 at 12:19
  • $\begingroup$ 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. $\endgroup$ – 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.

  • $\begingroup$ I'm a bit confused. The circle needs to pass through four points, two on the first hexagon and two on the second hexagon, with it containing one of the intersection points. However, this circle only passes through three points. $\endgroup$ – user181475 Oct 22 '14 at 10:04
  • $\begingroup$ It passes through two points on each hexagon including one common to both, which is what I thought you were asking. I too the four points to be the four mentioned. If you mean four different points you should find four which make a rectangle. $\endgroup$ – Mark Bennet Oct 22 '14 at 10:15
  • $\begingroup$ Oh, I meant that the circle passes through the four vertices, but the circle has one of the intersection points in its interior (not on the edge of the circle), one outside the circle. $\endgroup$ – user181475 Oct 22 '14 at 10:18
  • $\begingroup$ @user181475 I was going to delete this, but I think you need to be clearer in your question, and your comment here is one which needs to be visible to people who are trying to answer it. I am still not quite sure what you are asking. $\endgroup$ – Mark Bennet Oct 22 '14 at 10:25
  • $\begingroup$ I added some more notes in an effort to make it more clear. $\endgroup$ – user181475 Oct 23 '14 at 12:19

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