Maths Olympiad Prep

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Geometry Difficulty 4.6 AIME Prove it Italy

Problem:

Let ABCDEFA B C D E F be a hexagon inscribed in a circle such that AB=BCA B = B C, CD=DEC D = D E and EF=AFE F = A F. Prove that the segments ADA D, BEB E and CFC F are concurrent (that is, they have a point in common).

Solution

Solution:

Since AB=BCA B = B C, using the fact that in a circle congruent chords correspond to congruent inscribed angles, we have AEB=BEC\angle A E B = \angle B E C. Similarly CAD=DAE\angle C A D = \angle D A E and ACF=FCE\angle A C F = \angle F C E.

Hence ADA D, EBE B and CFC F are the three bisectors of triangle ACEA C E and therefore they are concurrent at its incenter.

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from it; metadata (topic, difficulty) added by this project.