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IMO Hk TST

Geometry Difficulty 7.6 National Olympiad, round 2 Prove it Hong Kong

ABCDEFABCDEF is a hexagon inscribed in a circle. Show that the diagonals ADAD, BEBE, CFCF are concurrent if and only if AB⋅CD⋅EF=BC⋅DE⋅FAAB \cdot CD \cdot EF = BC \cdot DE \cdot FA.

Solution

By Ceva's theorem, ADAD, BEBE, CFCF are concurrent if and only if
sin⁡∠CADsin⁡∠EAD×sin⁡∠AEBsin⁡∠CEB×sin⁡∠ECFsin⁡∠ACF=1. \frac{\sin \angle CAD}{\sin \angle EAD} \times \frac{\sin \angle AEB}{\sin \angle CEB} \times \frac{\sin \angle ECF}{\sin \angle ACF} = 1.
Figure 1
By the extended sine law, we have
sin⁡∠CADsin⁡∠EAD=CDDE,sin⁡∠AEBsin⁡∠CEB=ABBC,sin⁡∠ECFsin⁡∠ACF=EFFA. \frac{\sin \angle CAD}{\sin \angle EAD} = \frac{CD}{DE}, \quad \frac{\sin \angle AEB}{\sin \angle CEB} = \frac{AB}{BC}, \quad \frac{\sin \angle ECF}{\sin \angle ACF} = \frac{EF}{FA}.
Thus, ADAD, BEBE, CFCF are concurrent if and only if
CDDE×ABBC×EFFA=1. \frac{CD}{DE} \times \frac{AB}{BC} \times \frac{EF}{FA} = 1.
This is exactly AB⋅CD⋅EF=BC⋅DE⋅FAAB \cdot CD \cdot EF = BC \cdot DE \cdot FA.

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