Correct solutions were received from Bradley University community members
Alexander Nedyalkov, Dan Mikos, Bader Al-Kandari, Jeremy Light.
Further correct
solutions were
received from Dan Dima, Juan Carlos Carrara, Bill Webb, Juan Carlos Marivela,
Ahron Teitelman, Paul Botham, Jérôme Lefebvre, Ahmed AboAdham,
Paul Lee, Steve King, Iñigo Picaza, Nick McGrath, Seymur Cahangirov, Cee Ann
Franklin, Juan Carlos Marivela.
Use the notation of the diagram in the problem. Let A1,
A2, and A3 denote, respectively, the area of
the triangles AOB, BOC, and AOC. Then A1
+ A2 = A3 and
A1 = (OA)(OB) |sin(ÐAOB)|,
A2 = (OB)(OC) |sin(ÐBOC)|,
A3 = (OA)(OC) |sin(ÐAOC)|.
Since the lines lie at 60º to each
other, |sin(ÐAOB)|
= |sin(ÐBOC)|
= |sin(ÐAOC)|
= Ö3/2. Therefore (OA)(OB)
+ (OB)(OC) = (OA)(OC).
Dividing by (OA)(OB)(OC) gives perhaps the simplest
relationship 1/(OA)
-
1/(OB) + 1/(OC) = 1.
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ã2004 Alberto L. Delgado