數學知識交流---(2)南京2005

1) Since ∠ACO = 90, when points E and C overlap, AC is the tangent to semi-circle O.

Therefore with initial EC = OC - OE = 8 - 6 = 2 cm, we have when t = 1, the side CA of △ABC is tangential to semi-circle O.

2) When OE overlaps with CB (they can overlap since both are 12 cm, let F be the point of intersection between AB and the semi-circle O, i.e. F is on the semi-circular arc of the semi-circle.

With ∠BFC = 90 (∠ in semi-circle), we have:

CF = BC sin 30 = 6

BF = BC cos 30 = 6√3

Hence the overlapping area △BFC has an area = (1/2) x 6 x 6√3 = 18√3 cm2.