Van Khea's Quickie

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Problem

6Van Khea's Quickie, problem

Solution

$AI\,$ is the bisector of $\angle BAD,\,$ $AJ\,$ is the bisector of $\angle CAD.\,$ Since $\angle BAC=60^{\circ},\,$ $\angle IAJ=30^{\circ}.\,$

Van Khea's Quickie, illustration

But $\angle IAJ\,$ is inscribed into circle $(K),\,$ implying that it is subtended by the arc $\overset{\frown}{AJ}=60^{\circ}.\,$ The corresponding central $\angle IKJ=60^{\circ},\,$ and, since $\Delta IKJ\,$ is isosceles $(IK=JK),\,$ it is also equilateral.

Acknowledgment

The above problem was posted by Van Khea (Cambodia) at the Olimpiada pe Şcoală (The School Yard Olympiad) and commented with the above solution by Marian Dincă (Romania).

 

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