Monday, March 29, 2010

Fox 266 - Solutions

Solutions of Fox 266 by Joe, Bleaug, Yu, and Giannno are similar. There is a different geometric solution by Binary Descartes, we should add it in the following days.

No words needed. (Hey y not? say somethin' :)

Yu:Given phi=30°, then m(BOC) = 150° which is 1/2 of reflex m(BAC).
.:B, O and C are on the same circle centred at A. Hence AB = AO.

Giannno: Extend CA such a way that AD=AC. Then m(DBC)=90° (since BA=DC/2 =AD=AC) and m(D)=30° and since m(BOC)=150° we get DBOC cyclic quadrilateral while A is the center of the circle. Hence AB=AO=AC radii of the same circle.
AD perpendicular to BO
m(OCB) + m(OBC) = 30°
m(ODA) = 60°
Quadrilateral DBAC is concyclic
m(BAD) = m(OCB)
m(DBC) = m(DAC) = 60° - m(BAD) = 60° - m(OCB)
m(DBO) = m(DBC) - m(OBC) = 30°
Triangles DBO and ABO are isosceles. ==> AB = AO.

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