MathDB

3

Part of 2005 Bundeswettbewerb Mathematik

Problems(2)

3A + 2B = 180° ==> a^2 + bc = c^2

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

3/1/2005
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.
trigonometrygeometry proposedgeometry
Similar triangles ACD, AEF, AMN

Source: German Mathematical Competition BWM 2005, 2nd round, problem

9/1/2005
Two circles k1k_1 and k2k_2 intersect at two points AA and BB. Some line through the point BB meets the circle k1k_1 at a point CC (apart from BB), and the circle k2k_2 at a point EE (apart from BB). Another line through the point BB meets the circle k1k_1 at a point DD (apart from BB), and the circle k2k_2 at a point FF (apart from BB). Assume that the point BB lies between the points CC and EE and between the points DD and FF. Finally, let MM and NN be the midpoints of the segments CECE and DFDF. Prove that the triangles ACDACD, AEFAEF and AMNAMN are similar to each other.
geometrygeometric transformationrotationratiogeometry proposed