olymmath

CMO 2006 Problem 5

  • Paper

The vertices of a right triangle ABCA B C inscribed in a circle divide the circumference into three arcs. The right angle is at AA, so that the opposite arc⁡BC\operatorname{arc} B C is a semicircle while arc⁡AB\operatorname{arc} A B and arc⁡AC\operatorname{arc} A C are supplementary. To each of the three arcs, we draw a tangent such that its point of tangency is the midpoint of that portion of the tangent intercepted by the extended lines ABA B and ACA C. More precisely, the point DD on arc⁡BC\operatorname{arc} B C is the midpoint of the segment joining the points D′D^{\prime} and D′′D^{\prime \prime} where the tangent at DD intersects the extended lines ABA B and ACA C. Similarly for EE on arc ACA C and FF on arc ABA B.

Prove that triangle DEFD E F is equilateral.

Dataset text NuminaMath olympiads. Quoted with source.

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