+109 In the diagram below, AB = AC = 1, and arc BC is centered at A with \angle BAC = 60^\circ. Point D is on \overline{AB}, and point E is on arc BC so that BD = DE, and arc BE (centered at D) is tangent to \overline{AC}. Point F is on \overline{DB}, and point G is on arc DE so that DF = FG, and arc DG (centered at F) is tangent to \overline{AE}. Compute the length DF.
+15143
With AB = AC = 1 and angle BAC = 60°, it is an equilateral triangle with side length 1. D lies on AB, E lies on the arc BC. The condition BD = DE is only met if E is identical to C and D is identical to A. F lies on DB, B lies on the arc DE. The condition DF = FG is only met if F is identical to B and G is identical to C.
Thus, the length DF = AB = 1.
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+15143
With AB = AC = 1 and angle BAC = 60°, it is an equilateral triangle with side length 1. D lies on AB, E lies on the arc BC. The condition BD = DE is only met if E is identical to C and D is identical to A. F lies on DB, B lies on the arc DE. The condition DF = FG is only met if F is identical to B and G is identical to C.
Thus, the length DF = AB = 1.
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