Van Khea's Quickie

What Is This About?

Created with GeoGebra


6Van Khea's Quickie, problem


$AI\,$ 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.


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).


Related material

  • Angle Trisectors on Circumcircle
  • Equilateral Triangles On Sides of a Parallelogram
  • Pompeiu's Theorem
  • Pairs of Areas in Equilateral Triangle
  • The Eutrigon Theorem
  • Equilateral Triangle in Equilateral Triangle
  • Seven Problems in Equilateral Triangle
  • Spiral Similarity Leads to Equilateral Triangle
  • Parallelogram and Four Equilateral Triangles
  • A Pedal Property in Equilateral Triangle
  • Miguel Ochoa's van Schooten Like Theorem
  • Two Conditions for a Triangle to Be Equilateral
  • Incircle in Equilateral Triangle
  • When Is Triangle Equilateral: Marian Dinca's Criterion
  • Barycenter of Cevian Triangle
  • Excircle in Equilateral Triangle
  • Converse Construction in Pompeiu's Theorem
  • Wonderful Trigonometry In Equilateral Triangle
  • 60o Angle And Importance of Being The Other End of a Diameter
  • One More Property of Equilateral Triangles
  • |Contact| |Up| |Front page| |Contents| |Geometry| |Store|

    Copyright © 1996-2017 Alexander Bogomolny


    Search by google: