## A British Mathematics Olympiad Problem and its dual

Theorem 1: If the perpendicular bisectors of the alternate sides AB, CD & EF of a (convex) hexagon ABCDEF circumscribed around a circle are concurrent at the centre of the circle, then the lines PY, XR & QZ connecting opposite tangential points are concurrent.

Theorem 2: If the angle bisectors of the alternate angles B, D & F of a (convex) hexagon ABCDEF inscribed in a circle are concurrent at the centre of the circle, then the lines AD, BE & CF connecting opposite vertices are concurrent.

Corollary 1: If the perpendicular bisectors of the alternate sides AB, CD & EF of a (convex) hexagon ABCDEF circumscribed around a circle are concurrent at the centre of the circle, then it has 3 pairs of equal adjacent angles (e.g. angle A = angle B, angle C = angle D , and angle E = angle F).

Corollary 2: If the angle bisectors of the alternate angles B, D & F of a (convex) hexagon ABCDEF inscribed in a circle are concurrent at the centre of the circle, then it has 3 pairs of equal adjacent sides (e.g. EF = FA, AB = BC, and CD = DF).

Can you explain why (prove) the results are true? If stuck, have a look at my paper at A dual to a BMO problem.

Clicking on the 'Show Diagonals' button, reveals another interesting corollary of the 1st result, namely, that the main diagonals are concurrent at the same point of concurrency as the lines PY, XR & QZ. In general, according to Brianchon's theorem the diagonals of any hexagon circumscribed around a conic are concurrent. So in this case, we have that the lines PY, XR & QZ are concurrent at the Brianchon point.

Challenge: Can you explain why (prove) the BMO-Brianchon corollary above is also true? If stuck, have a look at Michael Fox's neat proof at BMO-Brianchon Concurrency problem.

The first result can be generalized to a hexagon circumscribed around a conic with the midpoints of the alternate sides AB, CD & EF of a (convex) hexagon ABCDEF touching the conic. For a dynamic sketch illustrating this generalization go to BMO Conic General problem.

The second result can be generalized to a cyclic hexagon, which has its two products of alternate sides equal, as shown at BMO Dual General problem.