## 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*).

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A British Mathematics Olympiad Problem and its dual

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*.

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Michael de Villiers, 30 October 2010.