Japan Mathematical Olympiad

Question

Suppose the points A, B, C, D are located on the circumference of a circle in this order as indicated in the figure below. Suppose the angle formed by the line tangent to the circle at B and the line AB is 30^, and that formed by the line tangent to the circle at C and the line CD is 10^. Suppose further that the lines AB and DC are parallel and they are located on the opposite sides from each other with respect to the center of the circle. Determine the magnitude of the angle BDC. FIGURE_0

Accepted Answer

Since AB DC and since DCA and DBA are angles subtended by the same arc AD at the points C and B on the circle, we have BDC = DBA = DCA. Since the angle subtended by an arc AB at the point C on the circle equals the angle formed by the tangent line to the circle at B and the chord AB, which is 30^, we have ACB = 30^. Similarly, we have DBC = 10^. Therefore, the sum of the inner angles of the triangle BCD equals BDC + DCA + ACB + DBC = 2 BDC + 30^ + 10^, from which we get BDC = 180^ - (10^ + 30^)2 = 70^.