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3A + 2B = 180° ==> a^2 + bc = c^2

Source: Bundeswettbewerb Mathematik 2005, 1st Round, problem 3

March 1, 2005
trigonometrygeometry proposedgeometry

Problem Statement

Let ABCABC be a triangle with sides aa, bb, cc and (corresponding) angles AA, BB, CC. Prove that if 3A+2B=180∘3A + 2B = 180^{\circ}, then a2+bc=c2a^2+bc=c^2. Additional problem: Prove that the converse also holds, i. e. prove the following: Let ABCABC be an arbitrary triangle. Then, 3A+2B=180∘3A + 2B = 180^{\circ} if and only if a2+bc=c2a^2+bc=c^2. Similar problem: Let ABCABC be an arbitrary triangle. Then, 3A+2B=360∘3A + 2B = 360^{\circ} if and only if a2−bc=c2a^2-bc=c^2.
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