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Geometry Difficulty 4.2 AIME Prove it Japan

Put 6 points AA, BB, CC, DD, EE, FF on the circumference of a circle in this order in such a way that the following conditions are satisfied: arc ABAB and arc BCBC, arc CDCD and arc DEDE, and arc EFEF and arc FAFA have same lengths, respectively. Determine the value of the angle ∠BFD\angle BFD if ∠ACE=68∘\angle ACE = 68^\circ. (Caution: the following diagram may not be accurate.)

Figure 1

Solution

From the well-known property of a quadrilateral inscribed in a circle, we have ∠ACE+∠AFE=180∘\angle ACE + \angle AFE = 180^\circ, and therefore, ∠AFE=180∘−68∘=112∘\angle AFE = 180^\circ - 68^\circ = 112^\circ.

Since the arcs CDCD and DEDE have the same lengths, we have from the theorem on inscribed angles that ∠CFD=∠DFE\angle CFD = \angle DFE, from which we conclude that ∠CFD=12∠CFE\angle CFD = \frac{1}{2} \angle CFE.

Similarly, we have ∠BFC=12∠AFC\angle BFC = \frac{1}{2} \angle AFC.

Consequently,
∠BFD=∠BFC+∠CFD=12(∠AFC+∠CFE)=12∠AFE=56∘. \angle BFD = \angle BFC + \angle CFD = \frac{1}{2}(\angle AFC + \angle CFE) = \frac{1}{2} \angle AFE = 56^\circ.

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