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Japan Junior Mathematical Olympiad

Geometry Difficulty 4.2 AIME Prove it Japan

Put 6 points $A$ , $B$ , $C$ , $D$ , $E$ , $F$ on the circumference of a circle in this order in such a way that the following conditions are satisfied: arc $AB$ and arc $BC$ , arc $CD$ and arc $DE$ , and arc $EF$ and arc $FA$ have same lengths, respectively. Determine the value of the angle $\angle BFD$ if $\angle ACE = 68^\circ$ . (Caution: the following diagram may not be accurate.)

Solution

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

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

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

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

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.